TS EAMCET · Physics · Oscillations
The position of a particle executing simple harmonic motion is given by \(x(t)=2 \cos \left(\frac{\pi}{15} t-\frac{\pi}{2}\right)\), where \(x\) is in centimetre and \(t\) is in seconds. The time period of the lsinetic energy of the particle in seconds is
- A \(\pi\)
- B \(\frac{\pi}{15}\)
- C 15
- D 30
Answer & Solution
Correct Answer
(C) 15
Step-by-step Solution
Detailed explanation
Given, position in SHM, \(x=2 \cos \left(\frac{\pi}{15} t-\frac{\pi}{2}\right)\) On comparing it with, we get \[ x=A \cos \left(\omega t-\frac{\pi}{2}\right) \] Angular frequency, \(\omega=\frac{\pi}{15}\) \[ \frac{2 \pi}{T}=\frac{\pi}{15} \Rightarrow T=30 \] So, time period of…
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