AP EAMCET · PHYSICS · Oscillations
A particle executing SHM along a straight line has zero velocity at points \(A\) and \(B\) whose distance from \(O\) on the same line \(O A B\) are \(a\) and \(b\), respectively. If the velocity at the mid point between \(A\) and \(B\) is \(v\), then its time period is
- A \(\frac{\pi(b+a)}{v}\)
- B \(\pi\left(\frac{b-a}{v}\right)\)
- C \(\left(\frac{b+a}{2 v}\right)\)
- D \(\left(\frac{b-a}{2 v}\right)\)
Answer & Solution
Correct Answer
(B) \(\pi\left(\frac{b-a}{v}\right)\)
Step-by-step Solution
Detailed explanation
According to the question, \(\therefore\) Amplitude \(=\frac{\text { Distance travelled by the particles }}{2}(A \text { to } B)\) Amplitude of particles executing simple harmonic motion (SHM) along a straight line \(A B\) is \((a)=\frac{b-a}{2}\). Velocity of particle, \(v=\)…
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