KCET · Physics · Oscillations
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
- A \( 2 \operatorname{rad} \mathrm{s}^{-1} \)
- B \( 0.5 \operatorname{rad} \mathrm{s}^{-1} \)
- C \( 2 \Pi \mathrm{rad} \mathrm{s}^{-1} \)
- D \( 0.5 \Pi \mathrm{rad} \mathrm{s}^{-1} \)
Answer & Solution
Correct Answer
(A) \( 2 \operatorname{rad} \mathrm{s}^{-1} \)
Step-by-step Solution
Detailed explanation
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 \(2 \mathrm{rads}^{-1}\)
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