MHT CET · Physics · Oscillations
A particle executes simple harmonic motion with amplitude 'A' and period \({ }^{\prime} \mathrm{T}^{\prime}\). If it is half way between mean position and extreme position, then its speed at that point is
- A \(\frac{3 \pi \mathrm{A}}{\mathrm{T}}\)
- B \(\frac{\sqrt{3} \pi \mathrm{A}}{2 \mathrm{~T}}\)
- C \(\frac{\pi \mathrm{A}}{\mathrm{T}}\)
- D \(\frac{\sqrt{3} \pi A}{\mathrm{~T}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\sqrt{3} \pi A}{\mathrm{~T}}\)
Step-by-step Solution
Detailed explanation
\(v=\omega \sqrt{\left(A^{2}-x^{2}\right)}=\frac{2 \pi}{T} \sqrt{A^{2}-\frac{A^{2}}{4}}=\frac{\sqrt{3} A}{2} \times \frac{2 \pi}{T}\)
\(=\frac{\sqrt{3} \pi \mathrm{A}}{\mathrm{T}}\)
\(=\frac{\sqrt{3} \pi \mathrm{A}}{\mathrm{T}}\)
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