JEE Main 2021 (Online) 17th March Evening Shift | Simple Harmonic Motion Question 74 | Physics

A block of mass 1 kg attached to a spring is made to oscillate with an initial amplitude of 12 cm. After 2 minutes the amplitude decreases to 6 cm. Determine the value of the damping constant for this motion . (take ln 2 = 0.693)

0.69 $$\times$$ 10² kg s^$$-$$1

3.3 $$\times$$ 10² kg s^$$-$$1

1.16 $$\times$$ 10^$$-$$2 kg s^$$-$$1

5.7 $$\times$$ 10^$$-$$3 kg s^$$-$$1

JEE Main 2021 (Online) 17th March Morning Shift

MCQ (Single Correct Answer)

-1

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal ?

x = $${A \over 2}$$

x = $$\pm$$ A

x = $$\pm$$ $${A \over {\sqrt 2 }}$$

x = 0

JEE Main 2021 (Online) 16th March Evening Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500g, Decay constant = 20 g/s then how much time is required for the amplitude of the system to drop to drop to half of its initial value? (ln 2 = 0.693)

17.32 s

34.65 s

0.034 s

15.01 s

JEE Main 2021 (Online) 16th March Morning Shift

MCQ (Single Correct Answer)

-1

Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g/2, the time period of pendulum will be :

$$\sqrt 3 T$$

$$\sqrt {{2 \over 3}} T$$

$${T \over {\sqrt 3 }}$$

$$\sqrt {{3 \over 2}} T$$

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