Oscillations Test

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Oscillations Test - 63

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Question 1
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A solid cylinder is attached to a horizontal massless spring as shown in figure. If the cylinder rolls without slipping, the time period of oscillation of the cylinder is?

A
$$2\pi\sqrt{\frac{x}{g}}$$

B
$$2\pi\sqrt{\frac{2M}{3K}}$$

C
$$2\pi\sqrt{\frac{3M}{8K}}$$

D
$$2\pi\sqrt{\frac{3M}{2K}}$$

Solution

Let x be the extension in the spring,
Then, $$x=2R\theta$$
$$\therefore$$ Torque acting on the cylinder about point of contact,
$$T=F\times 2R=-Kx(2R)=-4R^2K\cdot \theta$$
$$\therefore T=2\pi\sqrt{\displaystyle \frac{I}{Ke}}=2\pi\sqrt{\displaystyle\frac{\displaystyle \frac{3}{2}MR^2}{4R^2K}}=2\pi\sqrt{\displaystyle \frac{3M}{8K}}$$.
Question 2
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The time taken by a particle performing S.H.M. to pass from point A to B where its velocities are same is $$2$$ seconds. After another $$2$$ seconds, it returns to B. Determine the time period of oscillation is (in seconds).

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

2×( time taken to go from a to b + the next time taken to return at b) = $$2\times(2+2)= 8 sec(ans)$$
Question 3
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The spring constants of three springs connected to a mass M are shown in figure. When mass oscillates, what are the effective spring constant and the time period of vibration?

A
$$4k, 2\pi\sqrt{\displaystyle\frac{M}{4k}}$$

B
$$3k, 2\pi\sqrt{\displaystyle\frac{M}{3}}k$$

C
$$2k, 2\pi\sqrt{\displaystyle\frac{M}{2k}}$$

D
None of these

Solution

In lower segment/zone, two springs, each of constant k, are in parallel.
$$\therefore$$ Equivalent spring constant $$=(k+k)=2k$$
Now, for the system, there is a spring (of constant $$2k$$) in upper segment and a spring (of constant $$2k$$) in lower segment.
They are in series
$$\therefore k_r=\displaystyle\frac{k_1k_2}{(k_1+k_2)}=\frac{(2k)(2k)}{(2k+2k)}$$
$$\therefore k_r=4k^2/4k=k$$ and
$$\therefore T=2\pi\sqrt{\displaystyle\frac{M}{k}}$$.
Question 4
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A particle of mass $$m$$ is located in a one dimensional field where potential energy is given by:
$$V(x)=A(1-\cos\ {px})$$ where $$A$$ and $$p$$ are constants.
The period of small oscillations of the particle is

A
$$2\pi\sqrt { \cfrac { m }{ (Ap) } } $$

B
$$2\pi \sqrt { \cfrac { m }{ \left( A{ p }^{ 2 } \right) } } $$

C
$$2\pi \sqrt { \cfrac { m }{ A } } $$

D
$$\cfrac { 1 }{ 2\pi } \sqrt { \cfrac { Ap }{ m } } $$

Solution

We are given that a particle of mass $$m$$ is located in a one dimensional potential field and the potential energy is given by: $$V(x)=A(1-\cos {px})$$
So, we can find the force experienced by the particle as
$$F=-\cfrac{dV}{dx}=-Ap\sin {px}$$

For small oscillations, we have, $$F\approx -A{p}^{2}x$$
Hence, the acceleration would be given by, $$\alpha=\cfrac{F}{m}=-\cfrac{A{p}^{2}}{m}x$$
Also we know that, $$\alpha=\cfrac{F}{m}=-{\omega}^{2}x$$
So, $$\omega=\sqrt {\cfrac{A{p}^{2}}{m}}$$
$$\Rightarrow T=\cfrac{2\pi}{\omega}=2\pi \sqrt {\cfrac{m}{A{p}^{2}}}$$
Question 5
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A particle moves with simple harmonic motion in a straight line. In first $$T s$$, after starting from rest it travels a distance $$a$$ and in next $$Ts$$, it travels $$2\ a$$ in same direction, then

A
Amplitude of motion is $$3\ a$$

B
Time-period of oscillation is $$8 t$$

C
Amplitude of motion is $$4\ a$$

D
Time-period of oscillation is $$6 t$$

Solution

Consider a general equation,
$$x=Acos\omega t$$
Apply for OP,
$$A-a=Acos\omega t$$

similarly,
$$A-(a+2a)=Acos2\omega t$$
[$$\therefore$$ time doubled]
from here,
$$cos\omega t=\frac { A-a }{ A }$$
$$cos2\omega t=\frac { A-3a }{ A } \\ cos2\theta =2{ cos }^{ 2 }\theta -1\\ { \left( \frac { A-a }{ A } \right) }^{ 2 }=\frac { A-30+A }{ 2A }$$

from here,
$$A=2a$$
So, amplitude of the motion is $$2a$$.
$$2a-a=2acos\omega t\\ \frac { \pi }{ 3 } =\omega t\\ \frac { \pi }{ 3 } =\frac { 2\pi }{ T } t \\ T=6t$$

Time period$$=6t$$
Question 6
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A uniform meter stick is suspended through a small pin hole at the 10 cm mark. Find the time period of small oscillation about the point of suspension.

A
2s

B
5s

C
3s

D
1.55 s

Solution
Question 7
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To and fro motion is an example of

A
Oscillatory motion

B
Periodic motion

C
Circulatory motion

D
Both (A) and (B).

Solution
Question 8
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A light uniform rod of Young's modulus Y, cross sectional area A, coefficient of linear expansion $$\alpha $$ and length $$l_0$$ is rigidly connected to the support at one end and the other end of the rod is connected to the string as shown in the figure. The temperature of the rod is increased by $$\triangle\theta$$ with supports remaining fixed. Initially, the spring is in natural length position. Spring force on the rod acts uniformly over the cross section during elongation of the rod. Find the net of elongation of the rod. (Assume thermal strain to be small)

A
$$\frac{Kl_0\alpha\triangle\theta}{(K+YA/l_0)}$$

B
$$\frac{YA\alpha\triangle\theta}{(K+YA/l_0)}$$

C
$$YA\alpha\triangle\theta$$

D
None of the above

Solution

When the temperature is increases by $$\triangle \theta $$ then will elongation in rod and compression in spring.
At increased temperature the length of rod $$U$$ will be
$$l= l_{o}+ \triangle l$$
$$l= l_{o}+ l_{o} \alpha \triangle \theta$$
If the rod is elongated by amount $$\triangle x$$ then the net compression in the rod $$= l- (l_{o}+\triangle x)$$
$$=l_{o}+ l_{o} \alpha \triangle \theta - l_{o}- \triangle x$$
$$= l_{o} \alpha \triangle \theta - \triangle x$$
$$\Rightarrow $$ Strain (in rod) $$=\dfrac{l_{o} \alpha \triangle \theta - \triangle x}{l}$$
If the force exerted by rad on spring
$$F_{r}$$ and force exerted log $$s$$ bring is $$F_{s}$$ at point $$B$$.
then,
$$F_{r}= F_{s}$$
Since,
the compression in string $$=$$ elongation in rod $$= \triangle x$$
$$\Rightarrow F_{s}= k \triangle x$$
and,
we know that in red the youngs modulus
$$y= \dfrac{Stream}{Strain}$$
in $$Y=\dfrac{Fr/A}{(l_{o} \angle \triangle \theta - \triangle x)/l}$$
Where,
$$A \rightarrow $$ cross sectional area of rod
$$\Rightarrow F_{r}= \dfrac{AY (l_{o} \alpha \triangle \theta - \triangle x)}{l}$$
Since, $$F_{r}= F_{s}$$
$$\Rightarrow \dfrac{AY(l_{o} \alpha \triangle \theta - \triangle x)}{l}= k \triangle x$$
$$\Rightarrow AY l_{o} \alpha \triangle \theta - AY \triangle x= k \triangle x l$$
$$\Rightarrow \triangle x (k l + AY)= AY l_{o} \alpha \triangle \theta $$
$$\triangle x= \dfrac{AY l_{o} \alpha \triangle \theta }{k (l_{o}+ \triangle l)+ AY}= \dfrac{A Y l_{o} \alpha \triangle \theta }{k l_{o} \left[ \left(2+ \dfrac{\triangle l}{l_{o}} \right)+ \dfrac{AY}{l_{o}} \right]}$$
Assuming thermal strain $$\left( \dfrac{\triangle l}{l_{o}} \right)$$ to be negligible
$$\triangle x=\dfrac{AY \alpha \triangle \theta }{k+ \dfrac{AY}{l_{o}}}$$
Question 9
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A uniform disc of radius R is pivoted at point O on its circumference. The time period of small oscillations about an axis passing through O and perpendicular to plane of disc will be

A
T = 2(3R/2g)

B
T = 2(R/g)

C
T = 2(2R/g)

D
T = 2(2R/3g)

Solution

The time period of oscillations of a compound pendulum is given by T = 2(I/mgd)
For a disc with its axis passing through its circumference is $$l=\dfrac{mR^2}{2}+mR^2=\dfrac{3mR^2}{2}$$
Substituting the values, we get T = 2(3R/2g)
Question 10
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An air chamber of volume V has a neck of cross-sectional area a into which a light ball of mass m just fits and can move up and down without friction. The diameter of the ball is equal to that of the neck of the chamber. The ball is pressed down a little and released. If the bulk modulus of air is B, the time period of the oscillation of the ball is

A
$$T = 2 \pi \sqrt{\frac{Ba^2}{mV}}$$

B
$$T = 2 \pi \sqrt{\frac{BV}{ma^2}}$$

C
$$T = 2 \pi \sqrt{\frac{mB}{Va^2}}$$

D
$$T = 2 \pi \sqrt{\frac{mV}{Ba^2}}$$

Solution

The situation is as shown in the figure. Let P be pressure of air in the chamber. When the ball is pressed down a distance x, the volume of air decreases from V to say $$V - \triangle V$$. Hence the pressure increases from $$P to P + \triangle P$$ . The change in volume is, $$\triangle V = ax$$
The excess pressure $$\triangle P$$ is related to the bulk modulus B as
$$ \triangle P = B \frac{\triangle V}{V}$$
Restoring force on ball = excess pressure \times cross-sectional area
or $$F = \dfrac{Ba}{V} \triangle V$$
or $$F = -\dfrac{Ba^2}{V}x$$
or $$F = - kx$$, where $$k = \dfrac{Ba^2}{V}$$ i.e., $$F \propto -x$$
Hence, the motion of the ball is simple harmonic. If m is the ball, the time period of the SHM is
$$ T = 2 \pi \sqrt{\dfrac{m}{k}} or T = 2 \pi \sqrt{\dfrac{mV}{Ba^2}}$$