NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

Q. 13.1 Which of the following examples represent periodic motion?

(a) A swimmer completing one (return) trip from one bank of a river to the other and back.

(b) A freely suspended bar magnet displaced from its N-S direction and released.

(c) A hydrogen molecule rotating about its centre of mass.

(d) An arrow released from a bow

Answer:

(a) The motion is not periodic, though it is to and fro.

(b) The motion is periodic.

(c) The motion is periodic.

(d) The motion is not periodic.

Q. 13.2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

(a) the rotation of the earth about its axis.

(b) motion of an oscillating mercury column in a U-tube.

(c) motion of a ball bearing inside a smooth curved bowl when released from a point slightly above the lowermost point.

(d) general vibrations of a polyatomic molecule about its equilibrium position.

Answer:

(a) Periodic but not S.H.M.

(b) S.H.M.

(c) S.H.M.

(d) Periodic but not S.H.M.M [A polyatomic molecule has a number of natural frequencies, so its vibration is a superposition of SHM’s of a number of different frequencies. This is periodic but not SHM]

Q. 13.3 Fig. 13.18 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?

Fig 13.18

Answer:

The x-t plots for linear motion of a particle in Fig. 13.18 (b) and (d) represent periodic motion with both having a period of motion of two seconds.

Q. 13.4 (a) Which of the following functions of time represent

(a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give a period for each case of periodic motion ( $\omega$ is any positive constant):

(a) $sin\omega t-cos\omega t$

Answer:

$$\begin{gathered} sin\omega t-cos\omega t \\ =\sqrt{2}(\frac{1}{\sqrt{2}}sin\omega t-\frac{1}{\sqrt{2}}cos\omega t) \\ =\sqrt{2}(cos\frac{\pi }{4}sin\omega t-sin\frac{\pi }{4}cos\omega t) \\ =\sqrt{2}sin(\omega t-\frac{\pi }{4}) \end{gathered}$$

Since the above function is of form $Asin(\omega t+\varphi )$ it represents SHM with a time period of $\frac{2\pi }{\omega }$

Q. 13.4 (b) Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion ( $\omega$ is any positive constant):

(b) $si{n}^3\omega t$

Answer:

$$\begin{gathered} sin3\omega t=3sin\omega t-4si{n}^3\omega t \\ si{n}^3\omega t=\frac{1}{4}(3sin\omega t-sin3\omega t) \end{gathered}$$

The two functions individually represent SHM but their superposition does not give rise to SHM the motion will definitely be periodic with a period of $\frac{2\pi }{\omega }$

Q. 13.4 (c) Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give a period for each case of periodic motion ( $\omega$ is any positive constant):

(c) $3cos(\pi /4-2\omega t)$

Answer:

The function represents SHM with a period of $\frac{\pi }{\omega }$

Q.13.4 (d) Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant):

(d) $cos\omega t+cos3\omega t+cos5\omega t$

Answer:

Here, each individual function is SHM. But superposition is not SHM. The function represents periodic motion but not SHM.

$period=LCM(\frac{2\pi }{\omega },\frac{2\pi }{3\omega },\frac{2\pi }{5\omega })=\frac{2\pi }{\omega }$

Q. 13.4 (e) Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant):

(e) $exp(-{\omega }^2{t}^2)$

Answer:

The given function is exponential and, therefore, does not represent periodic motion.

Q. 13.4 (f) Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give a period for each case of periodic motion (ω is any positive constant):

(f) $1+\omega t+{\omega }^2{t}^2$

Answer:

The given function does not represent periodic motion.

Q. 13.5 (a) A particle is in linear simple harmonic motion between two points, A and B, $10cm$ apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is

(a) at the end A,

Answer:

Velocity is zero. Force and acceleration are in the positive direction.

Q. 13.5 (b) A particle is in linear simple harmonic motion between two points, $A$ and $B$ , $10cm$ apart. Take the direction from A to $B$ as the positive direction and give the signs of velocity, acceleration and force on the particle when it is

(b) at the end $B$,

Answer:

Velocity is zero. Acceleration and force are negative.

Q. 13.5 (c) A particle is in linear simple harmonic motion between two points, $A$ and $B$, $10cm$ apart. Take the direction from $A$ to $B$ as the positive direction and give the signs of velocity, acceleration and force on the particle when it is

(c) at the mid-point of AB going towards $A$,

Answer:

Velocity is negative, that is, towards A, and its magnitude is maximum. Acceleration and force are zero.

Q. 13.5 (d) A particle is in linear simple harmonic motion between two points, $A$ and $B$ , $10cm$ apart. Take the direction from $A$ to $B$ as the positive direction and give the signs of velocity, acceleration and force on the particle when it is

(d) at $2cm$ away from $B$ going towards $A$ ,

Answer:

The velocity is negative. Acceleration and force are also negative.

Q.13.5 (e) A particle is in linear simple harmonic motion between two points, $A$ and $B$, $10cm$ apart. Take the direction from $A$ to $B$ as the positive direction and give the signs of velocity, acceleration and force on the particle when it is

(e) at $3cm$ away from $A$ going towards $B$ , and

Answer:

Velocity is positive. Acceleration and force are also positive.

Q. 13.5 (f) A particle is in linear simple harmonic motion between two points, $A$ and $B$, $10cm$ apart. Take the direction from $A$ to $B$ as the positive direction and give the signs of velocity, acceleration and force on the particle when it is

(f) at $4cm$ away from $B$ going towards $A.$

Answer:

Velocity, acceleration and force all are negative

Q. 13.6 Which of the following relationships between the acceleration a and the displacement $x$ of a particle involve simple harmonic motion?

(a) $a=0.7x$

(b) $a=-200x^2$

(c) $a=-10x$

(d) $a=100x^3$

Answer:

Only the relation given in (c) represents simple harmonic motion as the acceleration is proportional in magnitude to the displacement from the midpoint, and its direction is opposite to that of the displacement from the mean position.

Q. 13.7 The motion of a particle executing a simple harmonic motion is described by the displacement function, $x(t)=Acos(\omega t+\varphi ).$

If the initial $(t=0)$ position of the particle is $1cm$ and its initial velocity is $\omega cm/s,$ what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi s^{-1}.$ If instead of the cosine function, we choose the sine function to describe the SHM: $x=Bsin(\omega t+\alpha ),$ what are the amplitude and initial phase of the particle with the above initial conditions?

Answer:

$\omega =\pi rad{s}^{-1}$

$x(t)=Acos(\pi t+\varphi )$

at t = 0

$$\begin{gathered} x(0)=Acos(\pi \times 0+\varphi ) \\ 1=Acos\varphi (i) \end{gathered}$$

$$\begin{gathered} v=\frac{\mathrm{d}x(t)}{\mathrm{d}t} \\ v(t)=-A\pi sin(\pi t+\varphi ) \end{gathered}$$

at t = 0

$$\begin{gathered} v(0)=-A\pi sin(\pi \times 0+\varphi ) \\ \omega =-A\pi sin\varphi \\ 1=-Asin\varphi (ii) \end{gathered}$$

Squaring and adding equation (i) and (ii), we get

$$\begin{gathered} {1}^2+1^2=(Acos\varphi {)}^2+(-Asin\varphi {)}^2 \\ 2=A^{2}co{s}^2\varphi +A^{2}si{n}^2\varphi \\ 2=A^2 \\ A=\sqrt{2} \end{gathered}$$

Dividing equation (ii) by (i) we get

$$\begin{gathered} tan\varphi =-1 \\ \varphi =\frac{3\pi }{4},\frac{7\pi }{4},\frac{11\pi }{4}...... \end{gathered}$$

$x(t)=Bsin(\pi t+\alpha )$

at t = 0

$$\begin{gathered} x(0)=Bsin(\pi \times 0+\alpha ) \\ 1=Bsin\alpha (iii) \end{gathered}$$

$$\begin{gathered} v=\frac{\mathrm{d}x(t)}{\mathrm{d}t} \\ v(t)=B\pi cos(\pi t+\alpha ) \end{gathered}$$

at t = 0

$$\begin{gathered} v(0)=B\pi cos(\pi \times 0+\alpha ) \\ \omega =B\pi cos\alpha \\ 1=Bcos\alpha (iv) \end{gathered}$$

Squaring and adding equation (iii) and (iv), we get

$$\begin{gathered} {1}^2+1^2=(Bsin\alpha {)}^2+(Bcos\alpha {)}^2 \\ 2=B^{2}si{n}^2\alpha +B^{2}co{s}^2\alpha \\ 2=B^2 \\ B=\sqrt{2} \end{gathered}$$

Dividing equation (iii) by (iv), we get

$$\begin{gathered} tan\alpha =1 \\ \alpha =\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4}...... \end{gathered}$$

Q. 13.8 A spring balance has a scale that reads from $0$ to $50kg.$ The length of the scale is $20cm,$ A body suspended from this balance, when displaced and released, oscillates with a period of $0.6s.$ What is the weight of the body?

Answer:

Spring constant of the spring is given by

$\begin{array}{rl}k& =\frac{\text{ Weight of Maximum mass the scale can read }}{\text{ Maximum displacement of the scale }}\\ k& =\frac{50\times 9.8}{20\times {10}^{-2}}\\ k& =2450{\mathrm{Nm}}^{-1}\end{array}$

The time period of a spring attached to a body of mass $m$ is given by

$\begin{array}{rl}T& =2\pi \sqrt{\frac{m}{k}}\\ m& =\frac{T^{2}k}{4{\pi }^2}\\ m& =\frac{(0.6{)}^2\times 2450}{4{\pi }^2}\\ m& =22.34\mathrm{kg}\\ w& =mg\\ w& =22.34\times 9.8\\ w& =218.95\mathrm{N}\end{array}$

Q. 13.9 (i) A spring having with a spring constant $1200N{m}^{-1}$ is mounted on a horizontal table as shown in Figure 13.19. A mass of $3kg$ is attached to the free end of the spring. The mass is then pulled sideways to a distance of $2.0cm$ and released.

Fig 13.19

Determine

(i) the frequency of oscillations,

Answer:

The frequency of oscillation of an object of mass m attached to a spring of spring constant k is given by

$\begin{array}{rl}\nu & =\frac{1}{2\pi }\sqrt{\frac{k}{m}}\\ \nu & =\frac{1}{2\pi }\times \sqrt{\frac{1200}{3}}\\ \nu & =3.183\mathrm{Hz}\end{array}$

Q. 13.9 (ii) A spring having with a spring constant $1200N{m}^{-1}$ is mounted on a horizontal table as shown in Figure 13.19. A mass of $3kg$ is attached to the free end of the spring. The mass is then pulled sideways to a distance of $2.0cm$ and released.

Fig 13.19

Determine

(ii) maximum acceleration of the mass, and

Answer:

A body executing S.H.M experiences maximum acceleration at the extreme points

$\begin{array}{rl}& a_{max}=\frac{F_A}{m}\\ & a_{max}=\frac{kA}{m}\\ & a_{max}=\frac{1200\times 0.2}{3}\\ & a_{max}=8{\mathrm{ms}}^{-2}\end{array}$

Q. 13.9 (iii) A spring having with a spring constant $1200N{m}^{-1}$ is mounted on a horizontal table as shown in Figure 13.9. A mass of $3kg$ is attached to the free end of the spring. The mass is then pulled sideways to a distance of $2.0cm$ and released.

Fig 13.19

Determine

(iii) the maximum speed of the mass.

Answer:

Maximum speed occurs at the mean position and is given by

$\begin{array}{rl}& v_{max}=A\omega \\ & v_{max}=0.02\times 2\pi \times 3.18\\ & v_{max}=0.4{\mathrm{ms}}^{-1}\end{array}$

Q. 13.10 (a) In Exercise 13.9, let us take the position of mass when the spring is unstretched as $x=0,$ and the direction from left to right as the positive direction of the x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch $(t=0),$ the mass is

(a) at the mean position,

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Answer:

Amplitude is A = 0.02 m

Time period is $\omega$

$\begin{array}{rl}\omega & =\sqrt{\frac{k}{m}}\\ \omega & =\sqrt{\frac{1200}{3}}\\ \omega & =20\mathrm{rad}/\mathrm{s}\end{array}$

(a) At t = 0 the mass is at mean position i.e. at t = 0, x = 0

$x(t)=0.02sin\left(20t\right)$

Here x is in metres and t is in seconds.

Q. 13.10 (b) In Exercise 13.9, let us take the position of mass when the spring is unstretched as $x=0,$ and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch $(t=0),$ the mass is

(b) at the maximum stretched position,

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Answer:

Amplitude is A = 0.02 m

Time period is $\omega$

$\begin{array}{rl}& \begin{array}{rl}\omega & =\sqrt{\frac{k}{m}}\\ \omega & =\sqrt{\frac{1200}{3}}\\ \omega & =20\mathrm{rad}/\mathrm{s}\end{array}\\ & \text{ (b) At }\mathrm{t}=0\text{ the mass is at the maximum stretched position. }\\ & \begin{array}{rl}& x(0)=\mathrm{A}\\ & \varphi =\frac{\pi }{2}\\ & x(t)=0.02\sin (20t+\frac{\pi }{2})\\ & x(t)=0.02\cos (20t)\end{array}\end{array}$

Here x is in metres and t is in seconds.

Q. 13.10 (c) In Exercise 13.9, let us take the position of mass when the spring is unstretched as $x=0,$ and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch $(t=0),$ the mass is

(c) at the maximum compressed position.

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Answer:

Amplitude is A = 0.02 m

Time period is $\omega$

$\begin{array}{rl}& \begin{array}{rl}\omega & =\sqrt{\frac{k}{m}}\\ \omega & =\sqrt{\frac{1200}{3}}\\ \omega & =20\mathrm{rad}/\mathrm{s}\end{array}\\ & \text{ (c) At }\mathrm{t}=0\text{ the mass is at the maximum compressed position. }\\ & \begin{array}{rl}& \mathrm{x}(0)=-\mathrm{A}\\ & \varphi =\frac{3\pi }{2}\\ & x(t)=0.02\sin (20t+\frac{3\pi }{2})\\ & x(t)=-0.02\cos (20t)\end{array}\end{array}$

Here x is in metres and t is in seconds.

The above functions differ only in the initial phase and not in amplitude or frequency.

Q. 13.11 Figures 13.20 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.

Fig 13.20

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.

Answer:

(a) Let the required function be $x(t)=a\sin (\pm \omega t+\varphi )$

Amplitude $=3\mathrm{cm}=0.03\mathrm{m}$
$\mathrm{T}=2\mathrm{s}$

$\begin{array}{rl}\omega & =\frac{2\pi }{T}\\ \omega & =\pi \mathrm{rad}\mathrm{s}\end{array}$

Since initial position $\mathrm{x}(\mathrm{t})=0,\varphi =0$
As the sense of revolution is clockwise

$\begin{array}{rl}& x(t)=0.03\sin (-\omega t)\\ & x(t)=-0.03\sin (\pi t)\end{array}$

Here x is in metres and t is in seconds.
(b) Let the required function be $x(t)=a\sin (\pm \omega t+\varphi )$

Amplitude $=2\mathrm{m}$

$\mathrm{T}=4\mathrm{s}$

$\begin{array}{rl}& \omega =\frac{2\pi }{T}\\ & \omega =\frac{\pi }{2}\mathrm{rad}\mathrm{s}\end{array}$

Since initial position $x(t)=-A,\varphi =\frac{3\pi }{2}$
As the sense of revolution is anti-clockwise

$\begin{array}{rl}& x(t)=2\sin (\omega t+\frac{3\pi }{2})\\ & x(t)=-2\cos \left(\frac{\pi }{2}t\right)\end{array}$

Here x is in metres and t is in seconds.

Q. 13.12 (a) Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial $(t=0)$ position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).

(a) $x=-2sin(3t+\pi /3)$

Answer:

$\begin{array}{rl}& x=-2\sin (3t+\pi /3)\\ & x=2\cos (3t+\frac{\pi }{3}+\frac{\pi }{2})\\ & x=2\cos (3t+\frac{5\pi }{6})\end{array}$

The initial position of the particle is $x(0)$

$\begin{array}{rl}& x(0)=2\cos (0+\frac{5\pi }{6})\\ & x(0)=2\cos \left(\frac{5\pi }{6}\right)\\ & x(0)=-\sqrt{3}\mathrm{cm}\end{array}$

The radius of the circle i.e. the amplitude is 2 cm
The angular speed of the rotating particle is $\omega =3\mathrm{rad}\mathrm{s}{}^{-1}$
Initial phase is

$\begin{array}{rl}\varphi & =\frac{5\pi }{6}\\ \varphi & ={150}^{\circ }\end{array}$

The reference circle for the given simple Harmonic motion is

Q. 13.12 (b) Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial $(t=0)$ position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).

(b) $x=cos(\pi /6-t)$

Answer:

$$\begin{gathered} x(t)=cos(\frac{\pi }{6}-t) \\ x(t)=cos(t-\frac{\pi }{6}) \end{gathered}$$

The initial position of the particle is x(0)

$$\begin{gathered} x(0)=cos(0-\frac{\pi }{6}) \\ x(0)=cos(\frac{\pi }{6}) \\ x(0)=\frac{\sqrt{3}}{2}cm \end{gathered}$$

The radius of the circle i.e. the amplitude is 1 cm

The angular speed of the rotating particle is $\omega =1rad{s}^{-1}$

Initial phase is

$$\begin{gathered} \varphi =-\frac{\pi }{6} \\ \varphi =-{30}^o \end{gathered}$$

The reference circle for the given simple Harmonic motion is

Q. 13.12 (c) Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial $(t=0)$ position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).

(c) $x=3sin(2\pi t+\pi /4)$

Answer:

$x=3sin(2\pi t+\pi /4)$

$$\begin{gathered} x=-3cos(2\pi t+\frac{\pi }{4}+\frac{\pi }{2}) \\ =3cos(2\pi t+\pi +\frac{\pi }{4}+\frac{\pi }{2}) \\ =3cos(2\pi t+\frac{3\pi }{2}+\frac{\pi }{4}) \\ =3cos(2\pi t+\frac{7\pi }{4}) \end{gathered}$$

At t= 0

$phase=\frac{7\pi }{4}$

The reference circle is as follows

Q. 13.12 (d) Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial $(t=0)$ position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).

(d) $x=2cos\pi t$

Answer:

$x(t)=2cos(\pi t)$

The initial position of the particle is x(0)

$$\begin{gathered} x(0)=2cos(0) \\ x(0)=2cm \end{gathered}$$

The radius of the circle i.e. the amplitude is 2 cm

The angular speed of the rotating particle is $\omega =\pi rad{s}^{-1}$

Initial phase is

$\varphi =0^o$

The reference circle for the given simple Harmonic motion is

Q. 13.13 (a) Figure 13.21 shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. $A$ force $F$ applied at the free end stretches the spring. Figure 13.21 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 13.21

(b) is stretched by the same force F.

Fig 13.21

(a) What is the maximum extension of the spring in the two cases?

Answer:

(a) Let us assume the maximum extension produced in the spring is x.

At maximum extension

$F=Kx$
$x=\frac{F}{k}$

(b) Let us assume the maximum extension produced in the spring is x. That is x/2 due to force towards left and x/2 due to force towards right

$F=k\frac{x}{2}+k\frac{x}{2}$
$\Rightarrow x=\frac{F}{k}$

Q. 13.14 The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of $1.0m$ If the piston moves with simple harmonic motion with an angular frequency of $200rad/min,$ what is its maximum speed?

Answer:

Amplitude of SHM = 0.5 m

angular frequency is

$\begin{array}{rl}& \omega =200\mathrm{rad}/\min \\ & \omega =3.33\mathrm{rad}/\mathrm{s}\end{array}$

If the equation of SHM is given by

$x(t)=A\sin (\omega t+\varphi )$

The velocity would be given by

$\begin{array}{rl}& v(t)=\frac{\mathrm{d}x(t)}{\mathrm{d}t}\\ & v(t)=\frac{\mathrm{d}(A\sin (\omega t+\varphi ))}{\mathrm{d}t}\\ & v(t)=A\omega \cos (\omega t+\varphi )\end{array}$

The maximum speed is therefore

$\begin{array}{rl}& v_{max}=A\omega \\ & v_{max}=0.5\times 3.33\\ & v_{max}=1.67{\mathrm{ms}}^{-1}\end{array}$

Q. 13.15 The acceleration due to gravity on the surface of moon is $1.7m{s}^{-2}$ What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is $3.5s$ ? (g on the surface of earth is 9.8 m s–2)

Answer:

The time period of a simple pendulum of length l executing S.H.M is given by

$\begin{array}{rl}& T=2\pi \sqrt{\frac{l}{g}}\\ & {\mathrm{g}}_{\mathrm{e}}=9.8\mathrm{m}{\mathrm{s}}^{-2}\\ & {\mathrm{g}}_{\mathrm{m}}=1.7\mathrm{m}{\mathrm{s}}^{-2}\end{array}$

The time period of the pendulum on the surface of Earth is ${\mathrm{T}}_{\mathrm{e}}=3.5\mathrm{s}$
The time period of the pendulum on the surface of the moon is $T_m$

$\begin{array}{rl}& \frac{T_m}{T_e}=\sqrt{\frac{g_e}{g_m}}\\ & T_m=T_e\times \sqrt{\frac{g_e}{g_m}}\\ & T_m=3.5\times \sqrt{\frac{9.8}{1.7}}\\ & T_m=8.4\mathrm{s}\end{array}$

Q .13.16 A simple pendulum of length l and having a bob of mass $M$ is suspended in a car. The car is moving on a circular track of radius $R$ with a uniform speed $V$. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

Answer:

Acceleration due to gravity = g (in downward direction)

Centripetal acceleration due to the circular movement of the car = a _c

$a_c=\frac{v^2}{R}$ (in the horizontal direction)

Effective acceleration is

$$\begin{gathered} {g}^{\prime }=\sqrt{g^2+a_c^2} \\ {g}^{\prime }=\sqrt{g^2+\frac{v^4}{R^2}} \end{gathered}$$

The time period is T'

$$\begin{gathered} {T}^{\prime }=2\pi \sqrt{\frac{l}{g^{\prime }}} \\ {T}^{\prime }=2\pi \sqrt{\frac{l}{\sqrt{g^2+\frac{v^4}{R^2}}}} \end{gathered}$$

Q. 13.17 A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ${\rho }_{ı}$ . The cork is depressed slightly and then released

Show that the cork oscillates up and down simply harmonically with a period $T=2\pi \sqrt{\frac{h\rho }{{\rho }_{1}g}}$ where $\rho$ is the density of the cork. (Ignore damping due to the viscosity of the liquid).

Answer:

Let the cork be displaced by a small distance x in downwards direction from its equilibrium position where it is floating.

The extra volume of fluid displaced by the cork is Ax

Taking the downwards direction as positive we have

$\begin{array}{rl}& \begin{array}{rl}& ma=-{\rho }_{1}gAx\\ & \rho Aha=-{\rho }_{1}gAx\\ & \frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}=-\frac{{\rho }_{1}g}{\rho h}x\end{array}\\ & \text{ Comparing with }\mathrm{a}=-\mathrm{kx}\text{ we have }\\ & \begin{array}{rl}k& =\frac{{\rho }_{1}g}{\rho h}\\ T& =\frac{2\pi }{\sqrt{k}}\\ T& =2\pi \sqrt{\frac{\rho h}{{\rho }_{1}g}}\end{array}\end{array}$

Q. 13.18 One end of a U-tube containing mercury is connected to a suction pump and the other end to the atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes a simple harmonic motion.

Answer:

Let the height of each mercury column be h.

The total length of mercury in both columns = 2h.

Let the cross-sectional area of the mercury column be A.

Let the density of mercury be $\rho$

When either of the mercury columns dips by a distance x, the total difference between the two columns becomes 2x.

The weight of this difference is $2Ax\rho g$

This weight drives the rest of the entire column to the original mean position.

Let the acceleration of the column be a. Since the force is restoring

$$\begin{gathered} 2hA\rho (-a)=2xA\rho g \\ a=-\frac{g}{h}x \end{gathered}$$

$\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}=-\frac{g}{h}x$ which is the equation of a body executing S.H.M

The time period of the oscillation would be

$T=2\pi \sqrt{\frac{h}{g}}$

Additional Questions

Q. 1 An air chamber of volume V has a neck area of cross section into which a ball of mass m just fits and can move up and down without any friction (Fig.1). Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Fig.1]

Fig 1

Answer:

Let the initial volume and pressure of the chamber be V and P.

Let the ball be pressed by a distance x.

This will change the volume by an amount ax.

Let the change in pressure be $\mathrm{\Delta }P$

Let the Bulk's modulus of air be K.

$K=\frac{\mathrm{\Delta }P}{\mathrm{\Delta }V/V}$
$\mathrm{\Delta }P=\frac{Kax}{V}$

This pressure variation would try to restore the position of the ball.

Since force is restoring in nature, displacement and acceleration due to the force would be in different directions.

$\begin{array}{rl}& F=a\mathrm{\Delta }P\\ & -m\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}=a\mathrm{\Delta }p\\ & \frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}=-\frac{k{a}^2}{mV}x\end{array}$

The above is the equation of a body executing S.H.M.

The time period of the oscillation would be

$T=\frac{2\pi }{a}\sqrt{\frac{mV}{k}}$

Q. 2 (a) You are riding in an automobile of mass $3000kg.$ . Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags $15cm$ when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by ${50}^o{/}_o$ during one complete oscillation. Estimate the values of

(a) the spring constant $K$

Answer:

Mass of automobile (m) = 3000 kg

There are a total of four springs.

Compression in each spring, x = 15 cm = 0.15 m

Let the spring constant of each spring be k

$4kx=mg$
$k=\frac{3000\times 9.8}{4\times 0.15}$
$k=4.9\times {10}^{4}N$

Q. 2 (b) You are riding in an automobile of mass $3000kg$ . Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags $15cm$ when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by ${50}^o{/}_o$ during one complete oscillation. Estimate the values of

(b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports $750kg.$ .

Answer:

The amplitude of oscillation decreases by 50 % in one oscillation i.e. in one time period.

$\begin{array}{rl}T& =2\pi \sqrt{\frac{m}{k}}\\ T& =2\pi \times \sqrt{\frac{3000}{4\times 4.9\times {10}^4}}\\ T& =0.77\mathrm{s}\end{array}$

For damping factor b we have

$\begin{array}{rl}& x=x_0{e}^{(-\frac{bt}{2m})}\\ & \mathrm{x}={\mathrm{x}}_0/2\\ & \mathrm{t}=0.77\mathrm{s}\\ & \mathrm{m}=750\mathrm{kg}\\ & e^{-\frac{0.75\mathrm{\%}}{2\times 750}}=0.5\\ & \ln \left(e^{-\frac{0.775}{2\times 780}}\right)=\ln 0.5\\ & \frac{0.77b}{1500}=\ln 2\\ & b=\frac{0.693\times 1500}{0.77}\\ & b=1350.2287\mathrm{kg}{\mathrm{s}}^{-1}\end{array}$

Q .4 A circular disc of mass $10kg$ is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be $1.5s.$ The radius of the disc is $15cm$ . Determine the torsional spring constant of the wire. (Torsional spring constant $a$ is defined by the relation $J=-a\theta$ , where J is the restoring couple and θ the angle of twist).

Answer:

$J=-a\theta$

Moment of Inertia of the disc about the axis passing through its centre and perpendicular to it is

$\begin{array}{rl}& I=\frac{M{R}^2}{2}\\ & J=I\frac{{\mathrm{d}}^2\theta }{\mathrm{d}{t}^2}\\ & -a\theta =\frac{M{R}^2}{2}\frac{{\mathrm{d}}^2\theta }{\mathrm{d}{t}^2}\\ & \frac{{\mathrm{d}}^2\theta }{\mathrm{d}{t}^2}=-\frac{2a}{M{R}^2}\theta \end{array}$

The period of Torsional oscillations would be

$\begin{array}{rl}& T=2\pi \sqrt{\frac{M{R}^2}{2a}}\\ & a=\frac{2{\pi }^{2}M{R}^2}{T^2}\\ & a=\frac{2{\pi }^2\times 10\times (0.15{)}^2}{(1.5{)}^2}\\ & a=1.97\mathrm{N}\mathrm{m}\mathrm{rad}{}^{-1}\end{array}$

Q. 5 (a) A body describes simple harmonic motion with an amplitude of $5cm$ and a period of $0.2s.$ Find the acceleration and velocity of the body when the displacement is

(a) $5cm$

Answer:

$\begin{array}{rl}& \mathrm{A}=5\mathrm{cm}=0.05\mathrm{m}\\ & \mathrm{T}=0.2\mathrm{s}\\ & \omega =\frac{2\pi }{T}\\ & \omega =\frac{2\pi }{0.2}\\ & \omega =10\pi \mathrm{rad}{\mathrm{s}}^{-1}\end{array}$

At displacement x acceleration is $a=-{\omega }^{2}x$

At displacement x velocity is $v=\omega \sqrt{A^2-x^2}$

(a)At displacement 5 cm

$\begin{array}{rl}& v=10\pi \sqrt{(0.05{)}^2-(0.05{)}^2}\\ & v=0\\ \\ & a=-(10\pi {)}^2\times 0.05\\ & a=-49.35m{s}^{-2}\end{array}$

Q. 5 (b) A body describes simple harmonic motion with an amplitude of $5cm$ and a period of $0.2s.$ Find the acceleration and velocity of the body when the displacement is

(b) $3cm$

Answer:

$\begin{array}{rl}& \mathrm{A}=5\mathrm{cm}=0.05\mathrm{m}\\ & \mathrm{T}=0.2\mathrm{s}\\ & \omega =\frac{2\pi }{T}\\ & \omega =\frac{2\pi }{0.2}\\ & \omega =10\pi \mathrm{rad}{\mathrm{s}}^{-1}\end{array}$

At displacement x acceleration is $a=-{\omega }^{2}x$

At displacement x velocity is $v=\omega \sqrt{A^2-x^2}$

(a)At displacement 3 cm

$\begin{array}{rl}& v=10\pi \sqrt{(0.05{)}^2-(0.03{)}^2}\\ & v=10\pi \sqrt{0.0016}\\ & v=10\pi \times 0.04\\ & v=1.257{\mathrm{ms}}^{-1}\\ \\ & a=-(10\pi {)}^2\times 0.03\\ & a=-29.61{\mathrm{ms}}^{-2}\end{array}$

Q. 5 (c) A body describes simple harmonic motion with an amplitude of $5cm$ and a period of $0.2s.$ Find the acceleration and velocity of the body when the displacement is

(c) $0cm$

Answer:

$\begin{array}{rl}& \mathrm{A}=5\mathrm{cm}=0.05\mathrm{m}\\ & \mathrm{T}=0.2\mathrm{s}\\ & \omega =\frac{2\pi }{T}\\ & \omega =\frac{2\pi }{0.2}\\ & \omega =10\pi \mathrm{rad}{\mathrm{s}}^{-1}\end{array}$

At displacement x acceleration is $a=-{\omega }^{2}x$

At displacement x velocity is $v=\omega \sqrt{A^2-x^2}$

(a)At displacement 0 cm

$\begin{array}{rl}& v=10\pi \sqrt{(0.05{)}^2-(0{)}^2}\\ & v=10\pi \times 0.05\\ & v=1.57{\mathrm{ms}}^{-1}\\ \\ & a=-(10\pi {)}^2\times 0\\ & a=0\end{array}$

Q. 6 A mass attached to a spring is free to oscillate, with angular velocity $\omega$ , in a horizontal plane without friction or damping. It is pulled to distance $x_0$ and pushed towards the centre with a velocity $v_0$ at time $t=0$ Determine the amplitude of the resulting oscillations in terms of the parameters $\omega$ , $x_0$ and $v_0$ . [Hint : Start with the equation $x=acos(\omega t+\theta )$ and note that the initial velocity is negative.]

Answer:

At the maximum extension of spring, the entire energy of the system would be stored as the potential energy of the spring.

Let the amplitude be A

$\begin{array}{rl}& \frac{1}{2}k{A}^2=\frac{1}{2}m{v}_0^2+\frac{1}{2}k{x}_0^2\\ & A=\sqrt{x_0^2+\frac{m}{k}{v}_0^2}\end{array}$