Simple Harmonic Motion Test

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Simple Harmonic Motion Test - 18

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Question 1
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A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

A
219 N

B
196 N

C
223 N

D
225 N

Solution

Hint: The reading of the scale can be used to find the spring constant.

Step 1: Find the spring constant.

Maximum mass that scale can read = M = 50 kg

Max-displacement of the spring = Length of the scale = l = 20cm = 0.2 m

Time period = T = 0.6 s

Maximum force exerted on spring = Mg = F

Spring constant,

Step 2: Find the mass of the block.

Time period is given by,

Weight of the body = mg = 22.36 x 9.8

= 219.12 N
Question 2
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The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?

A
50 m/min

B
150 m/min

C
100 m/min

D
120 m/min

Solution

Hint: The velocity of the piston will be maximum at the mean position.

Step 1: Find the amplitude.

The amplitude is 0.5 m.

Step 2: Find the maximum velocity.

The maximum velocity is given by,
v_max = ωA = 200 x 0.5 = 100 m/min
Question 3
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The acceleration due to gravity on the surface of the moon is 1.7 m s⁻². What is the time period of a simple pendulum on the surface of the moon if its time period on the surface of the earth is 3.5 s? (g on the surface of the earth is 9.8 m s⁻² )

A
7.3 s

B
6.4 s

C
5.5 s

D
8.4 s

Solution

Hint: The time period depends upon the effective acceleration due to gravity.
Question 4
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What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

A
100 s⁻¹

B
0

C
150s⁻¹

D
None of these

Solution

Hint: The acceleration of the pendulum falling freely will be equal to zero.

Explanation: When a simple pendulum mounted in a cabin falls freely under gravity, its acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.
Question 5
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A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

A

B

C

D

Solution

Hint: The time period depends on the effective acceleration of the bob.

Step 1: Find the net acceleration of the bob.
Question 6
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A cylindrical piece of cork of density ρ and base area A and height h floats in a liquid of density ρ_l. The cork is depressed slightly and then released. If the cork oscillates up and down simple harmonically then its period is:

A

B

C

D

Solution

Hint: By finding the restoring force on the cork, we can find the time period.

Step 1: Find the restoring force on the cork.

Let the cork is depressed by a distance x.

The volume of the water displaced by the cork = Ax

Upthrust on the cork = ρ_lAxg

Restoring force = - ρ_lAxg

Step 2: Find the time period of the cork.

Comparing it with F = −Kx

K = ρlAg

Mass of the cork = ρAh
Question 7
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An air chamber of volume V has a neck area of cross-section 'a' into which a ball of mass 'm' just fits and can move up and down without any friction (as shown in the figure). When the ball is pressed down a little and released, it executes SHM. The time period of oscillations is:

(assuming pressure-volume variations of air to be isothermal)

A

B

C

D

Solution

Hint: Bulk modulus of the air gives the change in pressure.

Step 1: Find the restoring force on the ball.

The bulk modulus of a gas is given by -
Question 8
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A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 sec. The radius of the disc is 15 cm. The torsional spring constant of the wire is:

A
2.1 Nm rad⁻¹

B
0.6 Nm rad⁻¹

C
3.2 Nm rad⁻¹

D
1.9 Nm rad⁻¹

Solution

Hint: The time period of the rotating disc is given by,

Mas of circular disc, M = 10kg

Radius of disc, r = 15cm = 0.15m

Time period = 1.5 s

Step 1: Find the moment of inertia of the disc.
Question 9
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A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 sec. What is the acceleration of the body when the displacement is 5 cm?

A
−3π² ms⁻²

B
5π² ms⁻²

C
−5π² ms⁻²

D
3π² ms⁻²

Solution

Hint: The acceleration is directly proportional to the displacement in SHM.

Step 1: Find the angular velocity.
Given y = 5cm = 0.05m
T = 0.2s
Question 10
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A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. What is the velocity of the body when the displacement is 3 cm?

A
0.4π m/s

B
0

C
0.5π m/s

D
0.3π m/s

Solution

Hint: The velocity of the particle in SHM is given by

Step: Use the formula of velocity.

When displacement is y, velocity is given by,