Oscillations Test

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Oscillations Test - 56

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Question 1
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A simple harmonic motion having an amplitude $$A$$ and time period $$T$$ is represented by the equation: $$y=5\sin{\pi\left(t+4\right)}$$. Then the values of $$A$$ (in $$m$$) and $$T$$ (in $$s$$) are :

A
$$A=5$$; $$T=2$$

B
$$A=10$$; $$T=1$$

C
$$A=5$$; $$T=1$$

D
$$A=10$$; $$T=2$$

Solution
Question 2
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Directions For Questions

A $$100\ g$$ block is connected to a horizontal massless spring of force constant $$25.6\ N/m$$. As shown in Fig $$4.151$$ (a), the block is free to oscillate on a horizontal frictionless surface. The block is displaced $$3\ cm$$ from the equilibrium position and at $$t=0$$, it is released from rest at $$x=0$$. It executes simple harmonic motion with the positive $$x-$$ direction indicated in Fig $$4.151$$ (a) The position-time $$(x-t)$$ graph of the motion of the block is as shown in Fig $$4.151$$

...view full instructions

(b) When the block is at position $$B$$ on the graph. its

A
position and velocity are positive

B
position is positive and velocity is negative

C
position in negative and velocity is positive

D
position and velocity are negative

Solution
Question 3
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Directions For Questions

A $$100\ g$$ block is connected to a horizontal massless spring of force constant $$25.6\ N/m$$. As shown in Fig $$4.151$$ (a), the block is free to oscillate on a horizontal frictionless surface. The block is displaced $$3\ cm$$ from the equilibrium position and at $$t=0$$, it is released from rest at $$x=0$$. It executes simple harmonic motion with the positive $$x-$$ direction indicated in Fig $$4.151$$ (a)
The position time $$(x-t)$$ graph of motion of the block is as shown in Fig $$4.151$$ (b)

...view full instructions

When the block is at position $$C$$ on the graph, its

A
velocity is maximum and acceleration is zero

B
velocity if minimum and acceleration is zero

C
velocity is zero and acceleration is negative

D
velocity is zero and acceleration if positive
Question 4
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Directions For Questions

A $$100\ g$$ block is connected to a horizontal massless spring of force constant $$25.6\ N/m$$. As shown in Fig $$4.151$$ (a), the block is free to oscillate on a horizontal frictionless surface. The block is displaced $$3\ cm$$ from the equilibrium position and at $$t=0$$, it is released from rest at $$x=0$$. It executes simple harmonic motion with the positive $$x-$$ direction indicated in Fig $$4.151$$ (a)The position-time $$(x-t)$$ graph of the motion of the block is as shown in Fig $$4.151$$

...view full instructions

(b) Position of the block as a function of time can now be expressed as

A
$$x=2\sqrt{3}\cos\left(16t+\dfrac{\pi}{6}\right)cm$$

B
$$x=3\cos\left(16t+\dfrac{\pi}{3}\right)cm$$

C
$$x=3.8\cos\left(16t+\dfrac{\pi}{6}\right)cm$$

D
$$x=3.2\cos \left(16t+\dfrac{\pi}{4}\right)cm$$

Solution

$$\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{25600}{100}}=16\ rad/s$$
Using energy conservation, $$\dfrac{1}{2}kx^{2}+\dfrac{1}{2}mv^{2}=\dfrac{1}{2}kA^{2}$$
$$A=\sqrt{x^{2}+\dfrac{m}{k}v^{2}}=\sqrt{9+3}=2\sqrt{3}$$
$$x=A\cos (\omega t+\phi)=2\sqrt{3}\cos (16t+\phi)$$
From boundary conditions $$t=0, \phi$$ can be obtained which is $$\pi/6$$
$$x=2\sqrt{3}\cos\left(16t+\dfrac{\pi}{6}\right)cm$$
Question 5
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A cork floating on the pond water executes a simple harmonic motion, moving up and down over a range of $$4\ cm$$. The time period of the motion is $$1\ s$$. At $$t=0$$, the cork is at its lowest position of oscillation, the position and velocity of the cork at $$t=10.5\ s$$, would be

A
$$2\ cm$$ above the mean position, $$0\ m/s$$

B
$$2\ cm$$ below the mean position, $$0\ m/s$$

C
$$1\ cm$$ above the mean position, $$2\sqrt 3 \pi \ m/s$$ up

D
$$1\ cm$$ below the mean position, $$2\sqrt 3 \pi \ m/s$$ up

Solution

As the range of motion is $$4\ cm$$, the amplitude of motion is $$+2\ cm$$. $$10.5\ s$$ is equal to $$10.5$$ time period of simple harmonic motion; so we have to find the height and position of cork after one-half of time period as $$\dfrac{T}{2}=0.5\ s$$. As at $$t=10\ s$$, the particle is at its lowest position, after half a time period the cork would be at its maximum height and velocity of cork at extreme position is zero.
Question 6
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Directions For Questions

A $$100\ g$$ block is connected to a horizontal massless spring of force constant $$25.6\ N/m$$. As shown in Fig. (a), the block is free to oscillate on a horizontal frictionless surface. The block is displaced $$3\ cm$$ from the equilibrium position and at $$t=0$$, it is released from rest at $$x=0$$. It executes simple harmonic motion with the positive $$x-$$ direction indicated in Fig. (a)
The position-time $$(x-t)$$ graph of the motion of the block is as shown in Fig.

...view full instructions

(b) Velocity of the block as a function of time can be expressed as

A
$$v=-48\sin \left(16t\dfrac{\pi}{2}\right)cm/s$$

B
$$v=-48\sin \left(16t\dfrac{\pi}{3}\right)cm/s$$

C
$$v=-56\sin \left(16t\dfrac{\pi}{4}\right)cm/s$$

D
$$v=-56\sin \left(16t\dfrac{\pi}{6}\right)cm/s$$
Question 7
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The displacement of a particle is represented by the equation
$$y = 3 \cos \left [ \dfrac{\pi }{4} - 2\omega t \right ]$$ the motion of the particle is

A
Simple harmonic with period $$\dfrac{2\pi }{\omega } $$

B
Simple harmonic with period $$\dfrac{\pi }{\omega } $$

C
Periodic but not simple harmonic.

D
Non-periodic.

Solution

As we know,
A Simple harmonic is produced when a force( called restoring force ) proportional to the displacement acts on a particle.

All sine and cosine functions of t are simple harmonic in nature.
Hence the motion is simple harmonic motion.

A simple harmonic motion is always periodic.
Now,
given equation is
$$ y=3cos[\dfrac{\pi}{4}-2\omega t]$$
on comparing with $$y=A_0cos(kx-\omega ' t)$$
we get,
$$\omega '=2\omega$$

$$\dfrac{2\pi}{T}=2\omega$$

$$T=\dfrac{\pi}{\omega}$$

Hence the motion is simple harmonic with time period $$\dfrac{\pi }{\omega } $$.
Verifies the option (b).
Question 8
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The displacement of a particle is represented by the equation $$y = \sin ^{3}\omega t$$ The motion is

A
non-periodic.

B
Periodic but not simple harmonic.

C
Simple harmonic with period $$\dfrac{2\pi }{\omega} $$

D
Simple harmonic with period $$\dfrac{\pi }{\omega} $$

Solution

As we know that all function of single sine and cosine function of "t" is an SHM.
But, here equation is:
$$y=sin^3\omega t=\dfrac{1}{4}[3sin\omega t-sin3\omega t]$$

As this motion is not represented by a single harmonic function, hence it is not SHM. As this motion involves sine and cosine functions, hence it is periodic motion.
Question 9
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The relation acceleration and displacement of four particles are given below.
Which one particle is executing simple harmonic motion?

A
$$a_{x} = + 2x$$

B
$$a_{x} = 2x^{2}$$

C
$$a_{x} = - 2x^{2}$$

D
$$a_{x} = - 2x$$

Solution

As we know,
In simple harmonic motion, force is proportional and opposite to displacement.
$$F=-kx$$
Applying the equation of motion
$$F = Ma,$$
$$a=-kx$$
Verifies the option (d).
Question 10
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Figure shows the circular motion of a particle. The radius of the circle, The period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is

A
$$x(t) = B \sin \left ( \dfrac{2 \pi t}{30} \right )$$

B
$$x(t) = B \cos \left ( \dfrac{\pi t}{15} \right )$$

C
$$x(t) = B \sin \left ( \dfrac{\pi t}{15} + \dfrac{\pi }{2} \right )$$

D
$$x(t) = B \cos \left ( \dfrac{\pi t}{15} + \dfrac{\pi }{2} \right )$$

Solution

$$ x(t)= A sin \dfrac {2\pi t}{T}$$ or $$ A cos \dfrac {2 \pi t}{T}$$

At t=0

$$x(t)=0 or A$$

So, the first option satisfies as the Simple harmonic equation.