JEE Main 2021 (Online) 22th July Evening Shift | Simple Harmonic Motion Question 68 | Physics

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 :

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

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

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

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

JEE Main 2021 (Online) 18th March Evening Shift

MCQ (Single Correct Answer)

-1

The function of time representing a simple harmonic motion with a period of $${\pi \over \omega }$$ is :

cos($$\omega$$t) + cos(2$$\omega$$t) + cos(3$$\omega$$t)

sin²($$\omega$$t)

sin($$\omega$$t) + cos($$\omega$$t)

3cos$$\left( {{\pi \over 4} - 2\omega t} \right)$$

JEE Main 2021 (Online) 18th March Morning Shift

MCQ (Single Correct Answer)

-1

The time period of a simple pendulum is given by $$T = 2\pi \sqrt {{l \over g}} $$. The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to be nearest integer is :-