Question

The displacement of a body executing simple harmonic motion is $y = 8\cos(\pi t)$ cm. The displacement of the body in the time interval between 1.5 s and 2.5 s is:

• A. 16 cm
• B. 8 cm
• C. 4 cm
• D. 0 cm

Hint:

In SHM, displacement repeats after every period $T = \dfrac{2\pi}{\omega}$. If time interval equals an integer multiple of $T/2$, displacement can be zero. Check $\cos(\pi t)$ carefully — the argument determines sign and phase. Always distinguish between net displacement and total distance traveled.

Answer 1

The Correct Option is D

• At $t_1 = 1.5$ s, $y_1 = 8\cos(1.5\pi) = 8 \times 0 = 0$.
• At $t_2 = 2.5$ s, $y_2 = 8\cos(2.5\pi) = 8 \times 0 = 0$.
• Hence, the displacement after one full time interval is zero since both positions correspond to the same equilibrium point.
• Therefore, the net displacement is 0 cm.