Question

The simple harmonic motion of a particle is given by \( x = a \sin 2\pi t \). Then the location of the particle from its mean position at time \( \frac{1}{8} \) s is

• A. $a$
• B. $\frac{a}{2}$
• C. $\frac{a}{\sqrt{2}}$
• D. $\frac{a}{4}$
• E. $\frac{a}{8}$

Hint:

Always convert time into angle using $\omega t$ before evaluating sine.

Answer 1

The Correct Option is C

Concept: SHM equation: \[ x = a \sin(\omega t) \]

Step 1: Given: \[ x = a \sin(2\pi t) \] So, \[ \omega = 2\pi \]

Step 2: Substitute $t = \frac{1}{8}$: \[ x = a \sin\left(2\pi \cdot \frac{1}{8}\right) \] \[ = a \sin\left(\frac{\pi}{4}\right) \]

Step 3: Use identity: \[ \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \]

Step 4: Final position: \[ x = \frac{a}{\sqrt{2}} \] Final Answer: \[ \frac{a}{\sqrt{2}} \]