Given, x=sin(πt+3π)
Velocity v=dtdx=πcos(πt+3π)
Now, at t=1s, v=πcos(π+3π)
v=2−π=2−3.14=−1.57ms−1⇒∣v∣=157cms−1
The equation of a particle executing simple harmonic motion is given by x=sinπ(t+31)m. At t=1s, the speed of particle will be (Given: π=3.14)
Held on 27 Jun 2022 · Verified 6 Jul 2026.
157cms−1
0cms−1
272cms−1
314cms−1
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