ExamBro
ExamBro
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

  1. A \(\pi\)
  2. B \(\frac{\pi}{15}\)
  3. C 15
  4. D 30
Verified Solution

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…
Same subject
Explore more questions on app
From TS EAMCET
Explore more questions on app