Oscillations Test

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Oscillations Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

The maximum velocity of a particle, executing simple harmonic motion with an amplitude $$7$$ mm, is $$4.4$$ m/s. The period of oscillation is :

A
$$100$$ s

B
$$0.01$$ s

C
$$10$$ s

D
$$0.1$$ s

Solution

$$v= \omega A =\frac{2\pi}{T}A$$
Take $$\pi \approx 22/7$$
Question 2
1 / -0

A particle is executing SHM with amplitude of 4 cm and has a maximum velocity of 10 cm/sec.
(a) At what displacement its velocity is 4 cm/sec?
(b) What is its velocity at displacement 2 cm?

A
$$\pm$$3.66cm, 8.66 cm

B
$$\pm$$4.66cm, 8.66 cm

C
$$\pm$$3.66cm, 9.66 cm

D
$$\pm$$5.66cm, 10.66 cm

Solution

(a) In SHM velocity at any position y is given by
v = $$\omega$$ $$\sqrt{A^{2} - y^{2}}$$
$$V_{max} = A \omega$$ at y = 0
10 = 4 $$\omega$$ $$\omega$$ = 5/2 rad/sec.
Now V = 4 cm/sec.at
4 = 5/2 $$\sqrt{4^{2} - y^{2}}$$
16 - $$y^{2}$$ = 2.56
$$y^{2}$$ = 13.46
y = $$\pm$$ $$\sqrt{13.46}$$ cm = $$\pm$$3.66cm
(b) At y = 2cm
v = 5/2 $$\sqrt{4^{2} - 2^{2}}$$ = 5/2 x 2$$\sqrt{3}$$ = 5 $$\sqrt{3}$$ = 8.66 cm
Question 3
1 / -0

If two springs of spring constant $$k_1$$ and $$k_2$$ are connected together in parallel, the effective spring constant will be

A
increase

B
decrease

C
remain same

D
none of the above

Solution

the effective spring constant = $$k_1+k_2$$

the correct option is (a)
Question 4
1 / -0

A small solid cylinder of mass M attached to a horizontal massless spring can roll without slipping along a horizontal surface. find its time period.

A
2$$\pi$$ $$\sqrt{M/2k}$$

B
$$2\pi$$ $$\sqrt{3M/5k}$$

C
2$$\pi$$ $$\sqrt{3M/2k}$$

D
$$\pi$$ $$\sqrt{3M/2k}$$

Solution

At any instant of rolling the cylinder has rotational and translational kinetic energy.
$$K_{rot}$$ = 1/2 l$$\omega^{2}$$ = 1/2 1/2M$$R^{2}$$ $$v^{2}$$/$$R^{2}$$ = 1/4 M$$v^{2}$$.
$$K_{trans}$$ = 1/2 $$Mv^{2}$$
$$U_{potential}$$ = 1/2 $$Kx^{2}$$
Total energy E = $$k_{rot}$$ + $$K_{trans}$$ + U
E = 3/4 $$Mv^{2}$$ + 1/2 $$Kx^{2}$$
Here DE/dt = 3/4M(2v) dV/dt $$\div$$ 1/2 k2x dx/dt = 0
Or, dv/dt = -2k 3M x
Hence $$\alpha$$ = $$-\omega^{2}$$x
Therefore the motion is simple harmonic
$$\omega$$ = 2k/3M
T = 2$$\pi$$ $$\sqrt{3M/2k}$$
Question 5
1 / -0

A body executing $$S.H.M.$$ along a straight line has a velocity of $$3\ ms^{-1}$$ when it is at a distance of $$4\ m$$ from its mean position and $$4\ ms^{-1}$$ when it is at a distance of $$3\ m$$ from its mean position. Its angular frequency and amplitude are:

A
$$2\ rad\ s^{-1}$$ & $$5\ m$$

B
$$1\ rad\ s^{-1}$$ & $$10\ m$$

C
$$2\ rad\ s^{-1}$$ & $$10\ m$$

D
$$1\ rad\ s^{-1}$$ & $$5\ m$$

Solution

Velocity $$v=\omega \sqrt{A^2-X^2}$$
$$3=\omega \sqrt{A^2-4^2}$$...(1)
$$4=\omega\sqrt{A^2-3^2}$$...(2)
$$\dfrac{3}{4} =\dfrac{\sqrt{A^2-4^2}}{\sqrt{A^2-3^2}}$$
$$\dfrac{9}{16} =\dfrac{A^2-16}{A^2-9}$$
$$16A^2-9A^2=16^2-9^2$$
$$7A^2=175$$
$$A^2=25$$
$$A=5$$
From (1), $$3=\omega \sqrt{5^2-4^2}$$
$$\omega=1 rad/sec$$
Question 6
1 / -0

Two particles $$A$$ and $$B$$ of equal masses are suspended from two massless springs of spring constants $$k_1$$ and $$k_2$$ respectively. If the maximum velocities, during oscillations, are equal, the ratio of amplitudes of $$A$$ and $$B$$ is :

A
$$\sqrt{\dfrac{k_1}{k_2}}$$

B
$$\dfrac{k_2}{k_1}$$

C
$$\sqrt{\dfrac{k_2}{k_1}}$$

D
$$\dfrac{k_1}{k_2}$$

Solution
Question 7
1 / -0

The acceleration displacement $$(a-x)$$ graph of a particle executing simple harmonic motion is shown in the figure. Find the frequency of oscillation.

A
$$\dfrac{1}{2\pi}\sqrt{\dfrac{\beta}{\alpha}}$$

B
$$\dfrac{1}{2\pi}\sqrt{\dfrac{\alpha}{\beta}}$$

C
$$\dfrac{1}{2\pi}\sqrt{\dfrac{\beta}{2\alpha}}$$

D
$$\dfrac{1}{2\pi}\sqrt{\dfrac{2\beta}{\alpha}}$$

Solution

Equation of given graph will be,
$$a=-\dfrac{\beta}{\alpha}X$$....(1)
$$a=-\omega^2X$$...(2)
Comparing both equations, $$\omega=\sqrt{\dfrac{\beta}{\alpha}}$$
So, frequency, $$f = \dfrac{\omega}{2\pi} = \dfrac{1}{2\pi}\sqrt{\dfrac{\beta}{\alpha}}$$
Question 8
1 / -0

A particle executes SHM with a time period of 12 s. Find the time taken by the particle to go directly from its mean position to half of its amplitude.

A
6

B
1

C
7

D
8

Solution

We know,
Equation of SHM,

$$y=A\sin\left(\dfrac{2\pi t}{T}\right)$$
Where, A=Amplitude,T=time period

At $$y=0$$ mean position,
$$0=A\sin\left(\dfrac{\pi t}{6}\right)$$
i.e $$t=0$$

At $$y=\dfrac{A}{2}$$
$$\dfrac{A}{2}=A\sin\left(\dfrac{\pi t}{6}\right)$$
i.e $$t=1s$$

Difference in times, is the time taken to reach from mean to half of the Amplitude,
i.e $$1\ sec$$

Option $$\textbf B$$ is the correct answer.
Question 9
1 / -0

A simple harmonic motion along the $$x$$-axis has the following properties: amplitude = $$0.5\ m$$, the time to go from one extreme position to other is, $$2\ s$$ and $$x = 0.3\ m$$ at $$t = 0.5\ s$$. The general equation of the simple harmonic motion is

A
$$x = \left( {0.5m} \right)\sin \left[ {\dfrac{{\pi t}}{2} + {8^ \circ }} \right]$$

B
$$x = \left( {0.5m} \right)\sin \left[ {\dfrac{{\pi t}}{2} - {8^ \circ }} \right]$$

C
$$x = \left( {0.5m} \right)\cos \left[ {\dfrac{{\pi t}}{2} + {8^ \circ }} \right]$$

D
$$x = \left( {0.5m} \right)\cos \left[ {\dfrac{{\pi t}}{2} - {8^ \circ }} \right]$$

Solution

$$x=A \sin(\omega t+\phi)$$
$$A=0.5$$
Half time period $$\dfrac{T}{2}=2sec$$
$$T=4sec$$
$$\omega=\dfrac{2\pi}{T}=\dfrac{\pi}{2}$$
$$x=0.5\sin(\dfrac{\pi}{2}t+\phi)$$....(1)
At $$t=0.5 sec,x=0.3$$ putting in (1),
$$0.3=0.5\sin(\dfrac{\pi}{2}\times 0.5+\phi)$$
$$\dfrac{3}{5}=\sin(\dfrac{\pi}{4}+\phi)=\sin(45+\phi)$$
$$45+\phi=37^\circ$$
$$\phi=-8^\circ$$
So, $$x=(0.5m)\sin(\dfrac{\pi}{2}t-8^\circ)$$
Question 10
1 / -0

The elastic potential energy of a stretched spring is given by $$E=50{x}^{2}$$ where $$x$$ is the displacement in meter and $$E$$ is in joule, then the force constant of the spring is

A
$$100J{m}^{-1}$$

B
$$100Nm$$

C
$$100J/{m}^{2}$$

D
$$50Nm$$

Solution

Let the force constant of the spring be $$k$$.
Elastic potential energy of a spring is given by $$E$$ =$$\dfrac{1}{2}kx^{2}$$. ....(i)
Given , elastic potential energy of stretched spring is $$E$$ = $$50$$ $$x^{2}$$ ....(ii)
Comparing equations (i) and (ii) :
$$\dfrac{1}{2}kx^{2}$$ = $$50$$ $$x^{2}$$
We get $$k$$ = $$100$$ $$\dfrac{J}{m^{2}}$$ .

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Oscillations Test - 36

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Oscillations Test - 36

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Question 1

$$100$$ s

$$0.01$$ s

$$10$$ s

$$0.1$$ s

Solution

Question 2

$$\pm$$3.66cm, 8.66 cm

$$\pm$$4.66cm, 8.66 cm

$$\pm$$3.66cm, 9.66 cm

$$\pm$$5.66cm, 10.66 cm

Solution

Question 3

increase

decrease

remain same

none of the above

Solution

Question 4

2$$\pi$$ $$\sqrt{M/2k}$$

$$2\pi$$ $$\sqrt{3M/5k}$$

2$$\pi$$ $$\sqrt{3M/2k}$$

$$\pi$$ $$\sqrt{3M/2k}$$

Solution

Question 5

$$2\ rad\ s^{-1}$$ & $$5\ m$$

$$1\ rad\ s^{-1}$$ & $$10\ m$$

$$2\ rad\ s^{-1}$$ & $$10\ m$$

$$1\ rad\ s^{-1}$$ & $$5\ m$$

Solution

Question 6

$$\sqrt{\dfrac{k_1}{k_2}}$$

$$\dfrac{k_2}{k_1}$$

$$\sqrt{\dfrac{k_2}{k_1}}$$

$$\dfrac{k_1}{k_2}$$

Solution

Question 7

$$\dfrac{1}{2\pi}\sqrt{\dfrac{\beta}{\alpha}}$$

$$\dfrac{1}{2\pi}\sqrt{\dfrac{\alpha}{\beta}}$$

$$\dfrac{1}{2\pi}\sqrt{\dfrac{\beta}{2\alpha}}$$

$$\dfrac{1}{2\pi}\sqrt{\dfrac{2\beta}{\alpha}}$$

Solution

Question 8

6

1

7

8

Solution

Question 9

$$x = \left( {0.5m} \right)\sin \left[ {\dfrac{{\pi t}}{2} + {8^ \circ }} \right]$$

$$x = \left( {0.5m} \right)\sin \left[ {\dfrac{{\pi t}}{2} - {8^ \circ }} \right]$$

$$x = \left( {0.5m} \right)\cos \left[ {\dfrac{{\pi t}}{2} + {8^ \circ }} \right]$$

$$x = \left( {0.5m} \right)\cos \left[ {\dfrac{{\pi t}}{2} - {8^ \circ }} \right]$$

Solution

Question 10

$$100J{m}^{-1}$$

$$100Nm$$

$$100J/{m}^{2}$$

$$50Nm$$

Solution

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