An SHM is given by $x = 5\sin(4\pi t + \pi/4)$ cm. The displacement at $t = 0.25$ s is:
A$5$ cm
B$2.5\sqrt 2$ cm
C$-5/\sqrt 2$ cm
D$-5$ cm
Answer & Solution
Correct answer: C. $-5/\sqrt 2$ cm
At $t = 0.25$: argument = $4\pi(0.25) + \pi/4 = \pi + \pi/4 = 5\pi/4$. $\sin(5\pi/4) = -\sin(\pi/4) = -1/\sqrt 2$. So $x = 5 \times (-1/\sqrt 2) = -5/\sqrt 2$ cm.
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