Oscillations Test

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Oscillations Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

uniform rod of mass m and length I is suspended about its end. Time period is

A
$$2 \pi \sqrt { \cfrac { l } { g } }$$

B
$$2 \pi \sqrt { \cfrac { 2l } { g } }$$

C
$$2 \pi \sqrt { \cfrac { 2 l } { 3 g } }$$

D
$$2 \pi \sqrt { \cfrac { l } { 3 g } }$$
Question 2
1 / -0

The system shown in lying on a smooth horizontal surface. Mass $$m$$ is constrained to move along the line $$3$$. Springs $$1$$ and $$2$$ at right angles with respect to one another are symmetrically arranged with respect to $$3$$ which is an elastic cord. The force constants of $$1,2$$ and $$3$$ are $$k$$ each and they are all just taut in the condition shown. For small oscillations of $$m$$ along $$3$$, time period is

A
$$\pi \sqrt { \dfrac { m }{ 2k } } \left( 1+\sqrt { 2 } \right) $$

B
$$2\pi \sqrt { \dfrac { m }{ 2k } } \left( 1+\sqrt { 2 } \right) $$

C
$$\dfrac {3\pi}{2} \sqrt { \dfrac { m }{ 2k } } \left( 1+\sqrt { 2 } \right) $$

D
$$\pi \sqrt { \dfrac { 2m }{ k } } \left( 1+\sqrt { 2 } \right) $$

Solution
Question 3
1 / -0

A block of mass $$m$$ is suspended from one end of a light spring as shown. The origin $$O$$ is considered at distance equal to natural length of spring from ceiling and vertical downward direction as positive $$Y-axis.$$ When the system is in equilibrium a bullet of mass $$m / 3$$ moving in vertical upward direction with velocity $$v _ { 0 }$$ strikes the block and embeds into it. As a result, the block (with bullet embedded into it) moves up and starts oscillating.
Mark out the correct statements:

A
The block-bullet system performs SHM about
$$y = \frac { m g } { k } .$$

B
The block-bullet system performs oscillatory motion
but not SHM about $$y = \frac { m g } { k }.$$

C
The block-bullet system performs SHM about
$$y = \frac { 4 m g } { 3 k }.$$

D
The block-bullet system performs oscillatory motion
but not SHM about $$y = \frac { 4 m g } { 3 k }.$$

Solution
Question 4
1 / -0

The differential equation representing the SHM of a particule is $$\cfrac {{9d^2y}} {dt^2} +4Y=0.$$The time period of the particle is fiven by

A
$${\cfrac \pi 3}sec$$

B
$$ \pi sec$$

C
$${\cfrac {2\pi} 3}sec$$

D
$${3\pi }sec$$

Solution

The differential equation of S.H.M is
$$9\dfrac{d^2y}{dt^2}+4y=0$$
$$9a+4y=0$$ ($$a=\dfrac{d^2x}{dt^2}$$)
$$a=-\dfrac{4}{9}x$$
$$\omega ^2=\dfrac{4}{9}$$ ($$a=-\omega ^2 x$$)
$$\omega =\dfrac{2}{3}$$
$$\dfrac{2\pi}{T}=\dfrac{2}{3}$$
$$T=3\pi$$
The correct option is D.
Question 5
1 / -0

Two point masses of 3.0 kg and 6.0 kg are attached to opposite ends of horizontal sprig whose spring constant is 300 N$$m^{-1}$$ as shown in the figure. The natural vibration frequency of the system is approximately.

A
4 Hz

B
3 Hz

C
2 Hz

D
1 Hz

Solution
Question 6
1 / -0

Two masses $${m_1}$$ and $${m_2}$$ are joined by a spring as shown. The system is dropped to the ground from a certain height. The spring will be :-

A
Stretched when $${m_2} > {m_1}$$

B
compressed when $${m_2} < {m_1}$$

C
neither compressed nor stretched only when $${m_1} = {m_2}$$

D
neither compressed nor stretched regardless of the values of $${m_1}$$ and $${m_2}$$

Solution

Since, both the masses $$m_1$$ and $$m_2$$ will fall with same acceleration i.e. acceleration due to gravity 'g'.
And.
hence there will be no relative acceleration of masses $$m_1$$ and $$m_2$$ with respect to each other.
Therefore,
There will be neither compression nor expansion in spring regardless of value of masses $$m_1$$ and $$m_2$$.
Question 7
1 / -0

A horizontal force $$F$$ is applied to the lower end of a uniform thin rod of mass $$4kg$$ and length $$L = 50\ cm$$ as shown in the figure. The rod undergoes only translational motion along the smooth horizontal surface. If $$F = 60\ N$$ determine the angle $$\theta$$ for translation motion of the rod. $$(g =10\ m/s^{2})$$.

A
$$\tan^{-1} \left (\dfrac {2}{3}\right )$$

B
$$\tan^{-1} \left (\dfrac {1}{2}\right )$$

C
$$\sin^{-1} \left (\dfrac {1}{5}\right )$$

D
$$\sin^{-1} \left (\dfrac {1}{4}\right )$$

Solution
Question 8
1 / -0

Two particles are executing identical simple harmonic motions described by the equations, $$x_1=a\cos \left(\omega t+\left(\dfrac{\pi}{6}\right)\right)$$ and $$x_2=a\cos \left(\omega t+\dfrac{\pi}{3}\right)$$. The minimum interval of time between the particles crossing the respective mean position is?

A
$$\dfrac{\pi}{2\omega}$$

B
$$\dfrac{\pi}{3\omega}$$

C
$$\dfrac{\pi}{4\omega}$$

D
$$\dfrac{\pi}{6\omega}$$

Solution

Equations are $$x_1=a\cos \left(\omega t+\dfrac {\pi}{6}\right)$$
and $$x_2 =a\cos \left(\omega t+\dfrac {\pi}{6}\right)$$
The first will pass through the mean position when $$x_1 =0$$ i.e., for instants $$t$$ for which $$\left(\omega t+\dfrac {\pi}{6}\right)=\dfrac {n \pi}{2}$$, where $$n$$ is an integer.
The smallest value for $$t$$ is $$n=1 $$

$$ \omega t_1=(\pi /2)-(\pi /6)=\pi /3$$.
The second will pass through the mean position when $$x_2 =0$$ i.e., for instants $$t$$ for which $$\left (\omega t+\dfrac {\pi}{3}\right)=\dfrac {m\pi}{2}$$
where $$m$$ is an integer.

The smallest value for $$t$$ is $$m=1 $$

$$ \omega t_2 = (\pi /2)-(\pi /3)=\pi /6$$

The smallest interval between the instants $$x_1 =0$$ and $$x_2=0$$ therefore,

$$\omega (t_1 - t_2)=\left(\dfrac {\pi}{3}-\dfrac {\pi}{6}\right)=\dfrac {\pi}{6}\ \Rightarrow \ t_1 - t_2=\dfrac {\pi}{6\omega}$$
Question 9
1 / -0

The potential energy of a particle executing SHM along the x-axis is given by $$U=U_0-U_0\cos ax$$. What is the period of oscillation?

A
$$2\pi\sqrt{\dfrac{ma}{U_o}}$$

B
$$2\pi\sqrt{\dfrac{U_o}{ma}}$$

C
$$\dfrac{2\pi}{a}\sqrt{\dfrac{m}{U_o}}$$

D
$$2\pi\sqrt{\dfrac{m}{aU_o}}$$

Solution

The potential energy of a particle executing along the x-axis is given by,

$$U=U_0-U_0cosax$$

$$U=U_0(1-cosax)$$

Force, $$F=-\dfrac{dU}{dx}=-aU_0sinax$$. . . . . .(1)

for small $$x$$, $$sinax=ax$$

$$ma'=-a^2U_0x$$

acceleration, $$a'=\dfrac{a^2 U_0}{m}x$$

From the above equation,

$$\omega ^2=\dfrac{a^2U_0}{m}$$ ($$a'=-\omega ^2 x$$)

$$T=\dfrac{2\pi}{a}\sqrt{\dfrac{m}{U_0}}$$

The correct option is C.
Question 10
1 / -0

At some instant, velocity and acceleration of a particle of mass '$$m$$' in SHM are '$$a$$' and '$$b$$' respectively. If its angular frequency is $$\omega,$$ then its maximum K.E. is

A
$$\frac { 1 } { 2 } m \left[ a ^ { 2 } + \frac { b ^ { 2 } } { \omega } \right]$$

B
$$\frac { 1 } { 2 } m \left[ a ^ { 2 } - \frac { b ^ { 2 } } { \omega ^ { 2 } } \right]$$

C
$$\frac { 1 } { 2 } m \left[ a ^ { 2 } + \frac { b ^ { 2 } } { \omega ^ { 2 } } \right]$$

D
$$\frac { 1 } { 2 } m \left[ a ^ { 2 } - \frac { b ^ { 2 } } { \omega } \right]$$

Solution