JEE Main 2021 (Online) 25th February Morning Slot | Simple Harmonic Motion Question 56 | Physics

JEE Main 2021 (Online) 25th February Morning Slot

MCQ (Single Correct Answer)

-1

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :

16 m/s²

2$$\pi$$² ms^$$-$$2

$$\pi$$² ms^$$-$$2

9.8 ms^$$-$$2

JEE Main 2021 (Online) 24th February Evening Slot

MCQ (Single Correct Answer)

-1

In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant k, the frequency of oscillation of given body is :

JEE Main 2021 (Online) 24th February Evening Slot Physics - Simple Harmonic Motion Question 59 English

JEE Main 2021 (Online) 24th February Evening Slot Physics - Simple Harmonic Motion Question 59 English

$${1 \over {2\pi }}\sqrt {{{2k} \over {Mg\sin \alpha }}} $$

$${1 \over {2\pi }}\sqrt {{k \over {Mg\sin \alpha }}} $$

$${1 \over {2\pi }}\sqrt {{{2k} \over M}} $$

$${1 \over {2\pi }}\sqrt {{k \over {2M}}} $$

JEE Main 2021 (Online) 24th February Evening Slot

MCQ (Single Correct Answer)

-1

When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is :

circular

straight line

parabolic

elliptical

JEE Main 2020 (Online) 6th September Evening Slot

MCQ (Single Correct Answer)

-1

When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is described by
y(t) = y₀ sin² $$\omega $$t, where 'y' is measured from the lower end of unstretched spring. Then $$\omega $$ is:

$$\sqrt {{g \over {{y_0}}}} $$

$${1 \over 2}\sqrt {{g \over {{y_0}}}} $$

$$\sqrt {{{2g} \over {{y_0}}}} $$

$$\sqrt {{g \over {2{y_0}}}} $$

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