A body oscillate with SHM according to equation (SI unit), \( x = 10\cos(2\pi t + \pi/4) \) at t = 1.5 s, calculate the speed of the body.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 44\,\mathrm{m\,s^{-1}} \)
In simple harmonic motion \( x = A\cos(\omega t + \phi) \), and the velocity is the time derivative of the displacement. Here \( A = 10\,\mathrm{m} \) and \( \omega = 2\pi\,\mathrm{rad\,s^{-1}} \).
\( \displaystyle v = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi) = -20\pi\sin\left(2\pi t + \frac{\pi}{4}\right) \)
At \( t = 1.5\,\mathrm{s} \) the phase is
\( \displaystyle 2\pi(1.5) + \frac{\pi}{4} = 3\pi + \frac{\pi}{4} \;\Rightarrow\; \sin\left(3\pi + \frac{\pi}{4}\right) = -\sin\frac{\pi}{4} = -\frac{1}{\sqrt{2}} \)
\( \displaystyle v = -20\pi \times \left(-\frac{1}{\sqrt{2}}\right) = 10\sqrt{2}\,\pi \approx 44.4\,\mathrm{m\,s^{-1}} \)
The speed is the magnitude of the velocity, about \( 44\,\mathrm{m\,s^{-1}} \).