Question

A particle initially at the mean position is executing simple harmonic motion with an angular frequency \( \frac{\pi}{4} \, \text{rad/s} \). The ratio of the distances travelled by the particle in the first second and second second is:

• A. \( 2:1 \)
• B. \( 1:1 \)
• C. \( (1 + \sqrt{3}) : 1 \)
• D. \( (1 + \sqrt{2}) : 1 \)

Hint:

In SHM, use displacement \( x = A \sin(\omega t) \) and compute position at given times to find distances.

Answer 1

The Correct Option is D

In SHM, distance travelled in time \( t \) from mean is related to the sine function. Use the formula for displacement:
\[ x(t) = A \sin(\omega t) \]
At \( t = 1 \), \( x_1 = A \sin\left(\frac{\pi}{4} \cdot 1\right) = A \cdot \frac{1}{\sqrt{2}} \)
At \( t = 2 \), \( x_2 = A \sin\left(\frac{\pi}{4} \cdot 2\right) = A \cdot 1 \)
Hence distance in 1st second: \( A \cdot \frac{1}{\sqrt{2}} \)
Distance in 2nd second: \( A (1 - \frac{1}{\sqrt{2}}) \)
Ratio = \( \frac{\frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}} = (1 + \sqrt{2}) : 1 \)