Search any question & find its solution
Question:
Answered & Verified by Expert
A particle executing SHM has a maximum speed of \( 0.5 \mathrm{~ms}^{-1} \) and maximum acceleration of
\( 1.0 \mathrm{~ms}^{-2} \). The angular frequency of oscillation is
Options:
\( 1.0 \mathrm{~ms}^{-2} \). The angular frequency of oscillation is
Solution:
2921 Upvotes
Verified Answer
The correct answer is:
\( 2 \operatorname{rad} \mathrm{s}^{-1} \)
Given, maximum speed, $v_{\max }=0.5 m s^{-1} ;$ maximum acceleration,$a_{\max }=1.0 m s^{-1} ;$ angular frequency, $\omega=?$
Now, we know $v_{\max }=\omega A \rightarrow(1)$
where A is amplitude of oscillation.
Also, $a_{\max }=\omega^{2} A \rightarrow(2)$
Using Eqs. (1) and (2), we get
$\frac{v_{\max }}{a_{\max }}=\frac{\omega A}{\omega^{2} A}$
$\Rightarrow \frac{v_{\max }}{a_{\max }}=\frac{1}{\omega}$
$\Rightarrow \omega=\frac{a_{\max }}{v_{\max }}=\frac{1.0}{0.5}$
$\Rightarrow \omega=2 \operatorname{rad} s^{-1}$
Therefore, angular frequency of oscillation is $2 \mathrm{rads}^{-1}$
Now, we know $v_{\max }=\omega A \rightarrow(1)$
where A is amplitude of oscillation.
Also, $a_{\max }=\omega^{2} A \rightarrow(2)$
Using Eqs. (1) and (2), we get
$\frac{v_{\max }}{a_{\max }}=\frac{\omega A}{\omega^{2} A}$
$\Rightarrow \frac{v_{\max }}{a_{\max }}=\frac{1}{\omega}$
$\Rightarrow \omega=\frac{a_{\max }}{v_{\max }}=\frac{1.0}{0.5}$
$\Rightarrow \omega=2 \operatorname{rad} s^{-1}$
Therefore, angular frequency of oscillation is $2 \mathrm{rads}^{-1}$
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.