Search any question & find its solution
Question:
Answered & Verified by Expert
A particle executing simple harmonic motion has a maximum speed of $40 \mathrm{~ms}^{-1}$ and maximum acceleration of $60 \mathrm{~ms}^{-2}$. The period of oscillation is
Options:
Solution:
2685 Upvotes
Verified Answer
The correct answer is:
$\frac{4 \pi}{3} \mathrm{~s}$
Given, maximum speed of SHM is $40 \mathrm{~ms}^{-1}$.
$v_{\max }=a \omega=40 \mathrm{~ms}^{-1}$...(i)
Maximum acceleration of SHM is $60 \mathrm{~ms}^{-2}$.
As,
$$
a_{\max }=a \omega^2=60 \mathrm{~ms}^{-2}...(ii)
$$
Dividing Eq. (ii) by Eq. (i), we get
$$
\begin{array}{rlrl}
& & \frac{\omega^2}{\omega} & =\frac{60}{40} \\
\Rightarrow & & \omega & =\frac{3}{2} \mathrm{~s}^{-1} \\
\text { We know, } & \omega & =2 \pi \mathrm{v} \\
\Rightarrow & \frac{3}{2} & =2 \pi \frac{1}{T} \Rightarrow \frac{1}{T}=\frac{3}{4 \pi} \\
\Rightarrow & & T & =\frac{4 \pi}{3} \mathrm{~s}
\end{array}
$$
Hence, time period of the oscillation is $\frac{4 \pi}{3} \mathrm{~s}$.
$v_{\max }=a \omega=40 \mathrm{~ms}^{-1}$...(i)
Maximum acceleration of SHM is $60 \mathrm{~ms}^{-2}$.
As,
$$
a_{\max }=a \omega^2=60 \mathrm{~ms}^{-2}...(ii)
$$
Dividing Eq. (ii) by Eq. (i), we get
$$
\begin{array}{rlrl}
& & \frac{\omega^2}{\omega} & =\frac{60}{40} \\
\Rightarrow & & \omega & =\frac{3}{2} \mathrm{~s}^{-1} \\
\text { We know, } & \omega & =2 \pi \mathrm{v} \\
\Rightarrow & \frac{3}{2} & =2 \pi \frac{1}{T} \Rightarrow \frac{1}{T}=\frac{3}{4 \pi} \\
\Rightarrow & & T & =\frac{4 \pi}{3} \mathrm{~s}
\end{array}
$$
Hence, time period of the oscillation is $\frac{4 \pi}{3} \mathrm{~s}$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.