JEE Main 2021 (Online) 27th July Evening Shift | Simple Harmonic Motion Question 63 | Physics

An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $$t = {T \over 4}s$$ starting from mean position. Assume that the initial phase of the oscillation is zero.

0.62 J

6.2 $$\times$$ 10^$$-$$3 J

1.2 $$\times$$ 10³ J

6.2 $$\times$$ 10³ J

JEE Main 2021 (Online) 27th July Morning Shift

MCQ (Single Correct Answer)

-1

A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is $${{3E} \over 4}$$ then its displacement 'y' is given by :

y = a

$$y = {a \over {\sqrt 2 }}$$

$$y = {{a\sqrt 3 } \over 2}$$

$$y = {a \over 2}$$

JEE Main 2021 (Online) 25th July Evening Shift

MCQ (Single Correct Answer)

-1

In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.

$${1 \over 2}$$

$${3 \over 4}$$

$${1 \over 3}$$

$${1 \over 4}$$

JEE Main 2021 (Online) 22th July Evening Shift

MCQ (Single Correct Answer)

-1

T₀ is the time period of a simple pendulum at a place. if the length of the pendulum is reduced to $${1 \over {16}}$$ times of its initial value, the modified time period is :

4 T₀

$${1 \over {4}}$$ T₀

T₀

8$$\pi$$ T₀

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