Oscillations Test

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Oscillations Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

A body executing SHM has its velocity $$10 cm/s$$ and $$7 cm/s$$ when its displacements from mean position are $$3 cm$$ and $$4 cm$$ respectively. The length of path is

A
$$10 cm$$

B
$$9.5 cm$$

C
$$4 cm$$

D
$$11.36 cm$$

Solution

$$v=\omega \sqrt{A^{2}-x^{2}}$$
$$v_{1}=10 cm /sec$$
$$v_{2}=7 cm /sec$$
$$x_{1}=3 cm$$
$$x_{2}=4cm$$

$$v_{1}=\omega\sqrt{A^{2}-x^{2}_{1}}$$

$$v_{2}=\omega\sqrt{A^{2}-x^{2}_{2}}$$

$$\therefore \dfrac{10}{7}=\dfrac{\omega \sqrt{A^{2}-(3)^{2}}}{\omega \sqrt{A^{2}-(4)^{2}}}$$

$$\dfrac{100}{49}=\dfrac{A^{2}-9}{A^{2}-16}$$

$$51A^{2}=1600-(9\times 49)$$
$$A=4.76 cm$$
Length of the path is $$2A = 9.52$$ cm
Question 2
1 / -0

The minimum phase difference between two SHM's is:
$$ y_{1}=\sin \left(\dfrac{\pi}{6}\right) \sin(\omega t)+\sin\left(\dfrac{\pi }{3}\right) \cos(\omega t )$$

$$ y_{2}=\cos\left(\dfrac{\pi}{6}\right) \sin(\omega t)+\sin \left(\dfrac{\pi}{3}\right) \cos(\omega t)$$ is:

A
$$\dfrac{\pi}{3} $$

B
$$\dfrac{\pi}{6} $$

C
$$\dfrac{\pi}{12} $$

D
$$0$$

Solution

$$y_{1}=\sin(\omega t)\sin\left(\dfrac{\pi }{6}\right)+ \cos$$ $$\omega t \sin\dfrac{\pi }{3}$$

$$=\sin$$ $$\omega t \cos\left(\dfrac{\pi }{3}\right)+\cos$$ $$\omega t \sin\dfrac{\pi }{3}$$

$$=\sin\left(\omega t +\dfrac{\pi }{3}\right)$$

Similarly for the second equation is:
$$y_{2}=\sin\left(\omega t +\dfrac{\pi }{6}\right)$$

Thus phase difference between the two waves is
$$\dfrac{\pi }{3}-\dfrac{\pi }{6}=\dfrac{\pi }{6} $$
Question 3
1 / -0

A particle executes SHM in a straight line. The maximum speed of the particle during its motion is $$V_{max}$$. Then the average speed of the particle during the SHM is :

A
$$\dfrac{V_{m}}{\pi }$$

B
$$\dfrac{V_{m}}{2\pi }$$

C
$$\dfrac{2V_{m}}{\pi }$$

D
$$\dfrac{3V_{m}}{\pi }$$

Solution

$$Vm=A\omega$$
$$V avg=\dfrac{distance \ \ traveled }{time \ \ taken}$$
$$=\dfrac{A}{T}$$
$$T=\dfrac{2\pi}{\omega}$$
$$Vavg=\dfrac{4 A\times \omega}{2\pi} = \dfrac{2 A \times \omega}{\pi}$$
$$Vavg =\dfrac{2 V_m}{\pi}$$
Question 4
1 / -0

A 1 kg mass executes SHM with an amplitude 10 cm, it takes $$2\pi$$ seconds to go from one end to the other end. The magnitude of the force acting on it at any end is :

A
0.1 N

B
0.2 N

C
0.5 N

D
0.05 N

Solution

As $$ w = \cfrac{2\pi}{T} = 1 \ rad/sec $$ ; Amplitude $$A = 0.1 m$$
magnitude of force $$ = m \times w^{2}.A$$
$$= 0.1 N$$
Question 5
1 / -0

A body executes S.H.M. under the action of a force $$F$$$$_{1}$$ with a time period $$4/5$$ seconds. If the force is changed to $$F$$$$_{2}$$, it executes S.H.M. with a time period $$3/5$$ seconds. If both the forces $$F$$$$_{1}$$and $$F$$$$_{2}$$ act simultaneously in the same direction on the body, then its time period in seconds is:

A
$$12/25$$

B
$$24/25$$

C
$$25/24$$

D
$$25/12$$

Solution

$$F_{1}=m\omega^{2}x$$
$$T=\dfrac{2\pi}{\omega}$$
$$F=ma=$$ $$4m\pi^{2} x \dfrac{1}{T^{2}}$$
$$\therefore F_{1}=\dfrac{4m\pi^{2}}{T_{1}^{2}}x$$
$$F_{2}=\dfrac{4m \pi^{2}}{T^{2}_{2}}x$$
$$F_{1}+F_{2}=\dfrac{ 4m \pi^{2}x}{T^{2}} + \dfrac{4m \pi^{2}x}{T^{2}_{2}}$$
$$\dfrac{1}{T^{2}}=\dfrac{1}{T^{2}_{1}}+\dfrac{1}{T^{2}_{2}}$$
$$\therefore T=\dfrac{12}{25}$$
Question 6
1 / -0

A particle free to move along the x-axis has potential energy given by $$U(x)=k[1-exp\left ( -x^{2} \right )]$$ for$$-\infty \leq x\leq +\infty ,$$ where k is a positive constant of appropriate dimensions. Then:

A
at points away from the origin, the particle is in unstable equilibrium.

B
for any finite non-zero value of x, there is a force directed away from the origin.

C
if its total mechanical energy is k/2 it has its minimum kinetic energy at the origin.

D
for small displacements from x = 0, the motion is SHM.

Solution

$$exp \left ( -x^{2} \right ) = 1 - x^{2} + \cfrac{x^{4}}{2!} + ---$$
For small $$x$$ : $$exp \left ( -x^{2} \right ) = (1 - x^{2})$$
Thus $$U(x) = K [1-(1-x^{2})] = K x^{2}$$
$$ F = - \dfrac{dU}{dx} = - \dfrac{d(kx^{2})}{dx} = -2 K x$$
Thus, the motion is an SHM.
Question 7
1 / -0

A particle of mass $$m$$ is executing oscillations about the origin on the $$x$$-axis. It's potential energy is $$U(x)=k\left | x \right |^{3}$$, where $$k$$ is a positive constant. If the amplitude of oscillation is $$a$$, then its time period $$T$$ is:

A
proportional to $$\dfrac{1}{\sqrt{a}} $$

B
independent of $$a$$

C
proportional to $$\sqrt{a}$$

D
proportional to $$a^{3/2}$$

Solution

$$U=k |x|^3 \Rightarrow F=-\dfrac{dU}{dx}=-3k|x|^2 ...(i)$$
Also, for SHM $$x=a \sin \omega t$$ and $$\dfrac{d^2x}{dt^2}+\omega ^2x=0$$
$$\Rightarrow acceleration\ a=\dfrac{d^2x}{dt^2}=-\omega ^2x \Rightarrow F=ma$$
$$=m \dfrac{d^2x}{dt^2}= - m \omega ^2x ...(ii)$$
From equation $$(i)$$ & $$ (ii)$$ we get $$\omega =\sqrt{\dfrac{3kx}{m}} $$
$$ \Rightarrow T=\dfrac{2 \pi}{\omega}= 2 \pi \sqrt{\dfrac{m}{3kx}}= 2 \pi \sqrt{\dfrac{m}{3k(a \sin \omega t)}} \Rightarrow T \propto \dfrac{1}{\sqrt a}$$
Question 8
1 / -0

A body of mass 1/4 kg is in S.H.M and its displacement is given by the relation $$y= 0.05 sin(20t+\dfrac{\pi }{2})$$ m. If $$t$$ is in seconds, the maximum force acting on the particle is:

A
$$5$$ N

B
$$2.5$$ N

C
$$10$$ N

D
$$0.25$$ N

Solution

We know that
$$F= m\omega^{2}A$$
$$\omega = 20 rad / sec$$
$$A = 0.05 m$$
Thus
$$F= \dfrac{1}{4}\times 20\times 20\times \dfrac{1}{20}$$
$$=5 N $$
Question 9
1 / -0

Two particles are in S.H.M. along parallel straight lines with same amplitude and time period. If they cross each other in opposite directions at the midpoint of mean and extreme positions. Phase difference between them is:

A
$$30^{\circ}$$

B
$$ 120^{\circ}$$

C
$$150^{\circ}$$

D
$$180^{\circ}$$

Solution

Since $$x=A\sin { \left( \omega t+\phi \right) }$$
$$\\ \dfrac { A }{ 2 } =A\sin { \left( \omega t+\phi \right) } \\ \Rightarrow \sin { \left( \omega t+\phi \right) } =\frac { 1 }{ 2 } \\ \Rightarrow \omega t+\phi =30^{ \circ }or\quad 150^{ \circ }$$
So the one particle has phase of $$30^{ \circ }$$ and another has $$150^{ \circ }$$
So the phase difference is $$150^{ \circ }-30^{ \circ }=120^{ \circ }$$
Question 10
1 / -0

A disc of mass $$M$$ is attached to a horizontal massless spring of force constant $$K$$ so that it can roll without slipping along a horizontal surface. If the disc is pushed a little towards right and then released, its center of mass executes SHM with a period of

A
$$2\pi $$ $$\sqrt{\dfrac{M}{K}}$$

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

C
$$2\pi $$ $$\sqrt{\dfrac{M}{2K}}$$

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

Solution

Spring exerts a force equal to $$Kx$$ on the center of mass of the disc.
Friction acts on opposite direction to cause rolling without slipping.
$$Kx-f=Ma_{cm}$$
$$fR=\dfrac{1}{2}MR^2\alpha$$
Solving gives $$f=\dfrac{Kx}{3}$$
Hence restoring force acting on sphere=$$kx-f=\dfrac{2Kx}{3}$$
Hence $$\omega=\sqrt{\dfrac{2K}{3M}}$$
Hence time period of oscillation=$$\dfrac{2\pi}{\omega}=2\pi\sqrt{\dfrac{3M}{2K}}$$

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Oscillations Test - 16

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Oscillations Test - 16

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Question 1

$$10 cm$$

$$9.5 cm$$

$$4 cm$$

$$11.36 cm$$

Solution

Question 2

$$\dfrac{\pi}{3} $$

$$\dfrac{\pi}{6} $$

$$\dfrac{\pi}{12} $$

$$0$$

Solution

Question 3

$$\dfrac{V_{m}}{\pi }$$

$$\dfrac{V_{m}}{2\pi }$$

$$\dfrac{2V_{m}}{\pi }$$

$$\dfrac{3V_{m}}{\pi }$$

Solution

Question 4

0.1 N

0.2 N

0.5 N

0.05 N

Solution

Question 5

$$12/25$$

$$24/25$$

$$25/24$$

$$25/12$$

Solution

Question 6

at points away from the origin, the particle is in unstable equilibrium.

for any finite non-zero value of x, there is a force directed away from the origin.

if its total mechanical energy is k/2 it has its minimum kinetic energy at the origin.

for small displacements from x = 0, the motion is SHM.

Solution

Question 7

proportional to $$\dfrac{1}{\sqrt{a}} $$

independent of $$a$$

proportional to $$\sqrt{a}$$

proportional to $$a^{3/2}$$

Solution

Question 8

$$5$$ N

$$2.5$$ N

$$10$$ N

$$0.25$$ N

Solution

Question 9

$$30^{\circ}$$

$$ 120^{\circ}$$

$$150^{\circ}$$

$$180^{\circ}$$

Solution

Question 10

$$2\pi $$ $$\sqrt{\dfrac{M}{K}}$$

$$2\pi $$ $$\sqrt{\dfrac{3M}{K}}$$

$$2\pi $$ $$\sqrt{\dfrac{M}{2K}}$$

$$2\pi $$ $$\sqrt{\dfrac{3M}{2K}}$$

Solution

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