MHT CET · Physics · Oscillations
The displacement of a particle executing S.H.M. is \(x=\mathrm{a} \sin (\omega \mathrm{t}-\phi)\). Velocity of the particle at time \(t=\frac{\phi}{\omega}\) is \(\left(\cos 0^{\circ}=1\right)\)
- A \(\omega \cos \phi\)
- B \(\mathrm{a} \omega\)
- C \(\omega \cos 2 \phi\)
- D \(-\mathrm{a} \omega \cos 2 \phi\)
Answer & Solution
Correct Answer
(B) \(\mathrm{a} \omega\)
Step-by-step Solution
Detailed explanation
Velocity of the particle is given as:
\(
\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a} \omega \cos (\omega \mathrm{t}-\phi)
\)
\(\therefore\) Velocity of the particle at time \(\mathrm{t}=\phi / \omega\) is given as,
\(\mathrm{v} =\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a} \omega \cos \left(\left(\omega \times \frac{\phi}{\omega}\right)-\phi\right)=\mathrm{a} \omega \cos 0 \)
\( \therefore \mathrm{v} =\mathrm{a} \omega\)
\(
\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a} \omega \cos (\omega \mathrm{t}-\phi)
\)
\(\therefore\) Velocity of the particle at time \(\mathrm{t}=\phi / \omega\) is given as,
\(\mathrm{v} =\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{a} \omega \cos \left(\left(\omega \times \frac{\phi}{\omega}\right)-\phi\right)=\mathrm{a} \omega \cos 0 \)
\( \therefore \mathrm{v} =\mathrm{a} \omega\)
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