Oscillations Test

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Oscillations Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

If at any instant of time the displacement of a harmonic oscillator is $$0.02\ \text{m}$$ and its acceleration is $$2\ \text{ms}^{-2}$$, its angular frequency at that instant will be

A
$$0.1\ \text{rads}^{-1}$$

B
$$1\ \text{rads}^{-1}$$

C
$$10\ \text{rads}^{-1}$$

D
$$100\ \text{rads}^{-1}$$

Solution

At any instant, neglecting the direction, we have, $$a=\omega ^{2}x$$.
Substituting the given values, we get $$\omega=10 rad/s$$. Hence, option C is correct.
Question 2
1 / -0

The equation of displacement of a particle executing SHM is $$x=0.40 \cos(2000 t+18)$$. The frequency of the particle is

A
$$10^3\ \text{Hz}$$

B
$$20\ \text{Hz}$$

C
$$2\times 10^3\ \text{Hz}$$

D
$$\displaystyle \dfrac{10^3}{\pi}\ \text{Hz}$$

Solution

$$\omega = 2000$$ here
Thus, $$f=\dfrac{\omega}{2\pi}=\dfrac{1000}{\pi}\ \text{Hertz}$$
Question 3
1 / -0

Identical springs of spring constant $$K$$ are connected in series and parallel combinations. A mass $$m$$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

A
1:4

B
1:2

C
4:1

D
2:1

Solution

From Formula 1. in series combination, ,
$$k_{eq}=\frac{K\times K}{K+K}=\dfrac{K}{2}$$.
$$\therefore f_{1}=\dfrac{1}{2\pi }\sqrt{\dfrac{K}{2m}}$$ ...(1)
For parallel combination, $$k_{eq}=2K$$
$$\therefore f_{2}=\dfrac{1}{2\pi }\sqrt{\dfrac{2K}{m}}$$...(2)
Taking ratio, we get answer as$$ 1:4$$ Option A is correct.
Question 4
1 / -0

In SHM, select the wrong statement, where $${F}$$ is the force, $${a}$$ is the acceleration and $${v}$$ is the velocity of the particle in SHM.

A
$$\overset{\rightarrow}{F}\times \overset{\rightarrow}{v}$$ is a null vector

B
$$|\overset{\rightarrow}{F}\times \overset{\rightarrow}{a}|=0$$ (always)

C
$$\overset{\rightarrow}{F}\times \overset{\rightarrow}{a}< {0}$$

D
$$\overset{\rightarrow}{a}.\overset{\rightarrow}{x}<0$$

Solution

The velocity, acceleration and thus the force are along the direction of motion in the SHM.
So the cross product of Force with velocity or acceleration is always zero.
Thus $$A$$ and $$B$$ are correct.
Also, acceleration is always in opposite direction to the displacement vector. So dot-product of acceleration and displacement is always negative. Thus $$D$$ is also correct.
$$C$$ is the wrong statement as the cross product of force and acceleration is strictly zero, because, $$F=ma\Rightarrow ma\times a=0$$.
Question 5
1 / -0

The following figure shows three identical springs $$A, B, C$$. When a 4 kg weight is hung on $$A$$, it descends by $$1$$ cm. When a $$6$$ kg weight is hung on $$C$$, it will descend by

A
1.5 cm

B
3.0 cm

C
4.5 cm

D
6.0 cm

Solution

The spring constant can be found out by
$$k=\dfrac{F}{x}=\dfrac{4\times 10}{0.01}=4000N/m$$
Now, when two springs are connected in series, their equivalent constant is
$$\dfrac{1}{k_{eq}}=\dfrac{1}{4000}+\dfrac{1}{4000}$$
$$k_{eq}=2000N/m$$
Therefore, mass of 6 kg will descend by
$$x=\dfrac{6\times 10}{2000}=0.03m=3cm$$
Option B is thus correct.
Question 6
1 / -0

A particle of mass $$M$$ is executing oscillations about the origin on the x-axis. Its potential energy is $$|U|=K|x^2|$$ where $$K$$ is a positive constant. If the amplitude of oscillations is $$a$$, then its period $$T$$ is

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

B
independent of $$a$$

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

D
proportional to $$a^{1/2}$$

Solution

The restoring force is $$-m\omega^2x=-dU/dx=-2Kx\Rightarrow \omega=\sqrt{2K/m}$$ where $$\omega$$ is the angular frequency of SHM.
Time period $$T=2\pi/\omega=2\pi\sqrt{m/2K}$$
So Time period is independent of amplitude $$a$$.
Question 7
1 / -0

The equation of S.H.M. of a particle is $$a+4\pi^{2}x=0$$ where $$a$$ is the instantaneous linear acceleration at displacement $$x$$. The frequency of motion is

A
$$1\space Hz$$

B
$$4\pi\space Hz$$

C
$$\frac{1}{4}\space Hz$$

D
$$4\space Hz$$

Solution

Comparing $$a+\omega^{2}x=0\ with\ a+4\pi^2x=0$$
$$\omega^{2}= 4\pi^{2} \Rightarrow \omega=2\pi ,
2\pi n=2\pi , n=1\ Hz$$
Question 8
1 / -0

Which of the following quantities is non-zero at the mean position for a particle executing SHM?

A
force

B
acceleration

C
velocity

D
displacement

Solution

At mean position, displacement, $$x=0$$. Since for SHM, acceleration is directly proportional to $$x$$, so acceleration is also 0.
Hence $$F=ma$$ is also zero.
But velocity is not zero (it is zero at extreme positions)
Question 9
1 / -0

A particle is executing SHM along a straight line $$8\ \text{cm}$$ long. While passing through mean position its velocity is $$16\ \text{cms}^{-1}$$. Its time period will be:

A
$$0.0157\ \text{s}$$

B
$$0.157\ \text{s}$$

C
$$1.57\ \text{s}$$

D
$$15.7\ \text{s}$$

Solution

8 cm long line means amplitude is 4 cm (both sides of mean)
Velocity at mean position is the maximum velocity given by $$A\omega$$
Thus $$16=4\omega$$ or $$\omega = 4$$
Thus, $$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{4}=\dfrac{\pi}{2}=1.57\ \text{s}$$
Question 10
1 / -0

The time period of a particle executing SHM is $$\displaystyle \frac{2\pi}{\omega}$$ and its velocity at a distance $$b$$ from mean position is $$\sqrt{3}b\omega$$. Its amplitude is

A
b

B
2b

C
3b

D
4b

Solution

$$x=A\sin\omega t$$
$$v=A\omega \cos\omega t$$
Thus, when $$x=b,$$ we have $$\sin\omega t = \dfrac{b}{A}$$

and hence, $$\cos\omega t=\sqrt{1-\left(\dfrac{b}{A}\right)^2}$$

Thus, $$\sqrt 3 b\omega = A\omega \sqrt{1-\left(\dfrac{b}{A}\right)^2}$$

or $$3b^2 = A^2\left(1-\dfrac{b^2}{A^2}\right)$$

or $$3b^2=A^2-b^2$$
Thus, $$A^2=4b^2$$ or $$A=2b$$

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Oscillations Test - 20

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Oscillations Test - 20

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Question 1

$$0.1\ \text{rads}^{-1}$$

$$1\ \text{rads}^{-1}$$

$$10\ \text{rads}^{-1}$$

$$100\ \text{rads}^{-1}$$

Solution

Question 2

$$10^3\ \text{Hz}$$

$$20\ \text{Hz}$$

$$2\times 10^3\ \text{Hz}$$

$$\displaystyle \dfrac{10^3}{\pi}\ \text{Hz}$$

Solution

Question 3

1:4

1:2

4:1

2:1

Solution

Question 4

$$\overset{\rightarrow}{F}\times \overset{\rightarrow}{v}$$ is a null vector

$$|\overset{\rightarrow}{F}\times \overset{\rightarrow}{a}|=0$$ (always)

$$\overset{\rightarrow}{F}\times \overset{\rightarrow}{a}< {0}$$

$$\overset{\rightarrow}{a}.\overset{\rightarrow}{x}<0$$

Solution

Question 5

1.5 cm

3.0 cm

4.5 cm

6.0 cm

Solution

Question 6

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

independent of $$a$$

proportional to $$\sqrt{a}$$

proportional to $$a^{1/2}$$

Solution

Question 7

$$1\space Hz$$

$$4\pi\space Hz$$

$$\frac{1}{4}\space Hz$$

$$4\space Hz$$

Solution

Question 8

force

acceleration

velocity

displacement

Solution

Question 9

$$0.0157\ \text{s}$$

$$0.157\ \text{s}$$

$$1.57\ \text{s}$$

$$15.7\ \text{s}$$

Solution

Question 10

b

2b

3b

4b

Solution

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