A simple pendulum performs simple harmonic motion about x=0 with an amplitude a and time period T . The speed of the pendulum at x= a 2 will be

Correct answer: D. π a 3 ~T aligned v=(d y)/(d t)=A ω & =A ω 1- ^2 ω t & =ω A^2-y^2 aligned Here, y= a 2 v=ω a^2- a^2 4 =ω 3 a^2 4 = 2 π T a 3 2 = π a 3 ~T

NEET 2009

Exam: NEET
Year: 2009
Subject: Physics
Chapter: Oscillations

NEET 2009 · Physics · Oscillations

A simple pendulum performs simple harmonic motion about x=0 with an...

A simple pendulum performs simple harmonic motion about

\mathrm{x}=0

with an amplitude a and time period

T

. The speed of the pendulum at

\mathrm{x}=\frac{\mathrm{a}}{2}

will be

Options

A $\frac{\pi \mathrm{a} \sqrt{3}}{2 \mathrm{~T}}$
B $\frac{\pi \mathrm{a}}{\mathrm{T}}$
C $\frac{3 \pi^2 a}{T}$
D $\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}$

Correct answer

D. $\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}$

Step-by-step solution

\begin{aligned}v=\frac{d y}{d t}=A \omega \cos \omegat & =A \omega \sqrt{1-\sin ^2 \omega t}\\ & =\omega \sqrt{A^2-y^2} \end{aligned}

Here,

\mathrm{y}=\frac{\mathrm{a}}{2}

\therefore

\mathrm{v}=\omega \sqrt{\mathrm{a}^2-\frac{\mathrm{a}^2}{4}}=\omega \sqrt{\frac{3 \mathrm{a}^2}{4}}=\frac{2 \pi}{\mathrm{T}} \frac{\mathrm{a} \sqrt{3}}{2}=\frac{\pi \mathrm{a} \sqrt{3}}{\mathrm{~T}}

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