Question:
A particle executing simple harmonic motion with an instantaneous displacement
\[
x=A\sin^2\left(\omega t-\frac{\pi}{4}\right)
\]
The time period of oscillation of the particle is
A particle executing simple harmonic motion with an instantaneous displacement
\[
x=A\sin^2\left(\omega t-\frac{\pi}{4}\right)
\]
The time period of oscillation of the particle is
Show Hint
Whenever \(\sin^2\theta\) or \(\cos^2\theta\) appears in SHM, convert it using double-angle identities because the effective angular frequency becomes doubled.
Updated On: Jun 22, 2026
- \(\dfrac{2\pi}{\omega}\)
- \(\dfrac{\pi}{\omega}\)
- \(\dfrac{\pi}{2\omega}\)
- \(\dfrac{\omega}{2\pi}\)
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The Correct Option is B
Solution and Explanation
Step 1: Use the trigonometric identity.
Given, \[ x=A\sin^2\left(\omega t-\frac{\pi}{4}\right) \] Using the identity, \[ \sin^2\theta=\frac{1-\cos2\theta}{2} \] Therefore, \[ x=A\left[\frac{1-\cos\left(2\omega t-\frac{\pi}{2}\right)}{2}\right] \] \[ x=\frac{A}{2}\left[1-\cos\left(2\omega t-\frac{\pi}{2}\right)\right] \]
Step 2: Determine the angular frequency.
The displacement contains the term \[ \cos(2\omega t-\frac{\pi}{2}) \] Hence, the angular frequency of oscillation is \[ 2\omega \]
Step 3: Find the time period.
The time period is given by \[ T=\frac{2\pi}{\text{angular frequency}} \] Thus, \[ T=\frac{2\pi}{2\omega} \] \[ T=\frac{\pi}{\omega} \]
Step 4: Final conclusion.
Hence, the time period of oscillation is \[ \boxed{\frac{\pi}{\omega}} \]
Given, \[ x=A\sin^2\left(\omega t-\frac{\pi}{4}\right) \] Using the identity, \[ \sin^2\theta=\frac{1-\cos2\theta}{2} \] Therefore, \[ x=A\left[\frac{1-\cos\left(2\omega t-\frac{\pi}{2}\right)}{2}\right] \] \[ x=\frac{A}{2}\left[1-\cos\left(2\omega t-\frac{\pi}{2}\right)\right] \]
Step 2: Determine the angular frequency.
The displacement contains the term \[ \cos(2\omega t-\frac{\pi}{2}) \] Hence, the angular frequency of oscillation is \[ 2\omega \]
Step 3: Find the time period.
The time period is given by \[ T=\frac{2\pi}{\text{angular frequency}} \] Thus, \[ T=\frac{2\pi}{2\omega} \] \[ T=\frac{\pi}{\omega} \]
Step 4: Final conclusion.
Hence, the time period of oscillation is \[ \boxed{\frac{\pi}{\omega}} \]
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