JEE Main 2021 (Online) 27th July Morning Shift | Simple Harmonic Motion Question 65 | Physics

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

JEE Main 2021 (Online) 20th July Evening Shift

MCQ (Single Correct Answer)

-1

A particle is making simple harmonic motion along the X-axis. If at a distances x₁ and x₂ from the mean position the velocities of the particle are v₁ and v₂ respectively. The time period of its oscillation is given as :

$$T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 - v_2^2}}} $$

$$T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 + v_2^2}}} $$

$$T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 + v_2^2}}} $$

$$T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 - v_2^2}}} $$

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