The bob of a simple pendulum of length L is released at time $t{\rm{ }} = {\rm{ }}0$ from a position of small angular displacement. Its linear displacement at time t is given by

${\rm{X}} = {\rm{a cos}}\sqrt {\frac{{\rm{g}}}{{\rm{L}}}} \times {\rm{t}}$

$X = a\sin 2\pi \sqrt {\frac{L}{{\rm{g}}}} \times t$

$X = a\cos 2\pi \sqrt {\frac{{\rm{g}}}{L}} \times t$

${\rm{X}} = {\rm{a sin}}\sqrt {\frac{{\rm{g}}}{{\rm{L}}}} \times {\rm{t}}$

NEET UG PHYSICS MCQ: OSCILLATIONS

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The bob of a simple pendulum of length L is released at time

t\mathrm{ } = \mathrm{ }0

from a position of small angular displacement. Its linear displacement at time t is given by

View answer and explanation

Correct option: D

\mathrm{X} = \mathrm{a cos}\sqrt {\frac{\mathrm{g}}{\mathrm{L}}} \times \mathrm{t}

Option D (

\mathrm{X} = \mathrm{a cos}\sqrt {\frac{\mathrm{g}}{\mathrm{L}}} \times \mathrm{t}

) is correct because the bob is released from a position of small angular displacement, implying a cosine function.

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{f_1}

{f_2}

F=-k x^3

(k=

)

f

2 f

2 f.

y_1=10 \sin \left(10 \pi t+\frac{\pi}{3}\right)

y_2=12 \sin \left(8 \pi t+\frac{\pi}{4}\right)

t=0.5 \mathrm{~s}