Oscillations Test

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Oscillations Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

A particle of mass 1kg is moving in a S.H.M with an amplitude of 0.02m and a frequency of 60Hz. The maximum force acting on the particle is :

A
$$2.88 \times 10^{3}$$N

B
$$1.44 \times 10^{3}$$N

C
$$5.67 \times 10^{3}$$N

D
$$0.75 \times 10^{3}$$N

Solution

$$f = 60\ Hz$$
$$w = 2\pi f = 120\pi \ rad/sec$$
$$F = mw^{2}A$$
$$= 1 \ kg \times 14400 \times \pi ^{2} \times 0.02$$
$$= 2.88 \times 10^{3}N$$
Question 2
1 / -0

The circular motion of a particle with constant speed is:

A
periodic but not SHM

B
SHM but not periodic

C
periodic and SHM

D
nethier periodic nor SHM

Solution

HINT: Circular motion is not oscillatory motion.
Step 1: Explanation
In circular motion because in the circular motion the particle does not move to and fro about a mean position as it is required for the SHM.
Question 3
1 / -0

The amplitude of oscillation of a particle is 0.05 m.If its period is 1.57s. Then the velocity at the mean position is

A
0.1 m/s

B
0.2 m/s

C
0.3 m/s

D
0.5 m/s

Solution

$$A=0.05m$$
$$T=\dfrac{\pi }{2}sec ; w=4\ { rad}/{sec}$$
$$v_{max}=wA =0.2 \ {m}/{s}$$
Question 4
1 / -0

A body is executing SHM. If the force acting on the body is 6N when the displacement is 2 cm, then the force acting on the body when the displacement is 3 cm in newton is:

A
$$ 6 $$ N

B
$$9$$ N

C
$$4$$ N

D
$$\sqrt{6} $$ N

Solution

$$ F = m w^{2}x$$
$$ \dfrac{F_{1}}{F_{2}} = \dfrac{x_{1}}{x_{2}}$$
$$ F_{2} = \dfrac{3}{2}\times 6 \ N = 9 \ N$$
Question 5
1 / -0

Water waves in a shallow dish are $$6.0\ m$$ long. At one point, the water moves up and down at a rate of $$4.8$$ oscillations/s. Speed of the wave is:

A
$$28.8\ m/s$$

B
$$26.8\ m/s$$

C
$$18.8\ m/s$$

D
$$38.8\ m/s$$
Question 6
1 / -0

A particle executes simple harmonic oscillation
with an amplitude a. The period of oscillation is
T. The minimum time taken by the particle to
travel half of the amplitude from the
equilibrium position is:

A
T/2

B
T/4

C
T/8

D
T/12

Solution

$$\because X=asin\omega t$$

$$\therefore \frac{a}{2}=aisn\omega t \Rightarrow \omega t=\frac{\pi }{6}$$

$$\Rightarrow \left ( \frac{2\pi }{T} \right )t=\frac{\pi }{6}\Rightarrow t=\frac{T}{12}$$
Question 7
1 / -0

The velocity of a particle that executes S.H.M. is at its mean position 0.866 m/s.Velocity at a displacement half of its amplitude from mean position is

A
1 m$$ s^{-1} $$

B
1.414m$$ s^{-1} $$

C
0.5m$$ s^{-1} $$

D
0.75m$$ s^{-1} $$

Solution

$$v_{max}=A\omega = 0.866\dfrac{m}{s}$$
At $$x=\dfrac{A}{2} ; v=\omega \sqrt{A^{2}-x^{2}}$$
$$v=\dfrac{\sqrt{3}}{2}\omega A=\dfrac{\sqrt{3}}{2}v_{max}$$
$$v=\dfrac{\sqrt{3}}{2}\times 0.866=0.75\dfrac{m}{s}$$
Question 8
1 / -0

If the displacement $$x$$ and velocity $$v$$ of a particle executing S.H.M are related through the expression $$ 4v^{2}=25-x^{2} $$, then its maximum displacement in meters is

A
1

B
2

C
5

D
6

Solution

$$v=\omega\sqrt{A^{2}-x^{2}}$$
$$4v^{2}=5^{2} -x^{2} \implies v=\frac{1}{2} \sqrt{(5^{2}-x^{2})}$$
Comparing the equation with the standard equation we get $$A = $$5m
Question 9
1 / -0

The displacement equations of two simple harmonic oscillators are given by $$ x_{1} =A_{1}\cos\omega t$$; $$ x_{2}= A_{2} \sin\left(\omega t+\frac{\pi }{6} \right )$$. The phase difference between them is:

A
$$ 30^{\circ} $$

B
$$ 60^{\circ} $$

C
$$ 90^{\circ} $$

D
$$ 120^{\circ} $$

Solution

$$X_{1}=A_{1} cos\omega t =A_{1} sin (\omega t+^{\pi}/_{2})$$
$$X_{2}=A_{2} sin (\omega t+^{\pi}/_{6})$$
so, phase difference $$=\dfrac{\pi}{2}-\dfrac{\pi}{6}=\dfrac{\pi}{3}=60^{0}$$
Question 10
1 / -0

Assertion (A): The phase difference between displacement and velocity in SHM is $$ 90^{\circ} $$
Reason (R): The displacement is represented by y=A sin$$ \omega $$t and Velocity by V=A$$ \omega $$cos$$ \omega $$t.

A
Both A and R are true and R is the correct explanation of A

B
Both A and R are true and R is not the correct explanation of A

C
A is true and R is false

D
A is false and R is true

Solution

At mean displacement is Zero but velocity is maximum At extreme position displacement is maximum but velocity is Zero.
Thus displacement can be represented by
$$y$$= $$A \ sin \ wt$$
$$V=\dfrac{dy}{dt}$$

$$=Awcoswt= Aw sin (wt+\dfrac{\pi}{2})$$
Thus phase difference is $$\dfrac{\pi}{2}$$

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Oscillations Test - 10

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Oscillations Test - 10

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Question 1

$$2.88 \times 10^{3}$$N

$$1.44 \times 10^{3}$$N

$$5.67 \times 10^{3}$$N

$$0.75 \times 10^{3}$$N

Solution

Question 2

periodic but not SHM

SHM but not periodic

periodic and SHM

nethier periodic nor SHM

Solution

Question 3

0.1 m/s

0.2 m/s

0.3 m/s

0.5 m/s

Solution

Question 4

$$ 6 $$ N

$$9$$ N

$$4$$ N

$$\sqrt{6} $$ N

Solution

Question 5

$$28.8\ m/s$$

$$26.8\ m/s$$

$$18.8\ m/s$$

$$38.8\ m/s$$

Question 6

T/2

T/4

T/8

T/12

Solution

Question 7

1 m$$ s^{-1} $$

1.414m$$ s^{-1} $$

0.5m$$ s^{-1} $$

0.75m$$ s^{-1} $$

Solution

Question 8

1

2

5

6

Solution

Question 9

$$ 30^{\circ} $$

$$ 60^{\circ} $$

$$ 90^{\circ} $$

$$ 120^{\circ} $$

Solution

Question 10

Both A and R are true and R is the correct explanation of A

Both A and R are true and R is not the correct explanation of A

A is true and R is false

A is false and R is true

Solution

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