Q132 / 160 · AP EAMCET 2022 (06 Jul Shift 1) · Physics
A particle executes simple harmonic motion with a time period $0.6 \mathrm{~s}$ and amplitude $10 \mathrm{~cm}$. Then the mean velocity of the particle over the time interval during which it travels a distance $5 \mathrm{~cm}$ starting from the equilibrium position.
- Correct answerA$1 \mathrm{~ms}^{-1}$
- B$50 \mathrm{~cm} \mathrm{~s}^{-1}$
- C$10 \mathrm{~cm} \mathrm{~s}^{-1}$
- D$1 \mathrm{~cm} \mathrm{~s}^{-1}$
Solution
We have
$$
\begin{aligned}
& x=A \sin \omega t \Rightarrow 5=10 \sin \omega t \\
& \Rightarrow \frac{1}{2}=\sin \omega t \Rightarrow t=\frac{\pi}{6 \omega}=\frac{\pi \times T}{6 \times 2 \pi}=\frac{T}{12}
\end{aligned}
$$
Then, $V=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{A} \omega \cos \omega \mathrm{t}$
$$
\Rightarrow \mathrm{V}=\mathrm{A} \omega \cos \omega \mathrm{t}
$$
Now,
$$
\begin{aligned}
& \mathrm{V}_{\text {mean }}=\frac{\int_0^{\mathrm{T} / 12} \mathrm{~A} \omega \cos \omega \mathrm{dt}}{\int_0^{\mathrm{T} / 12} \mathrm{dt}}=\frac{\frac{\mathrm{A} \omega}{\omega}[\sin \omega \mathrm{t}]_0^{\mathrm{T} / 12}}{\mathrm{~T} / 12} \\
& =\frac{\mathrm{A}\left[\frac{1}{2}-0\right]}{\frac{\mathrm{T}}{12}}=\frac{\frac{0.1}{2}}{\frac{0.6}{12}}=\frac{12}{12}=1 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.