Question

The displacement of a particle executing simple harmonic motion is given by \[ x=6\sin\left(2\pi t+\frac{\pi}{4}\right)\,\text{m}. \] The amplitude and maximum speed are respectively

• A. \(4\,\text{m},\,2\pi\,\text{m s}^{-1}\)
• B. \(6\,\text{m},\,4\pi\,\text{m s}^{-1}\)
• C. \(6\,\text{m},\,12\pi\,\text{m s}^{-1}\)
• D. \(2\,\text{m},\,12\pi\,\text{m s}^{-1}\)

Hint:

For SHM of the form \[ x=A\sin(\omega t+\phi), \] \[ A=\text{coefficient of sine or cosine}, \] and \[ v_{\max}=A\omega. \] The phase constant \(\phi\) does not affect the amplitude or maximum speed.

Answer 1

The Correct Option is C

Concept: \(x = 6\sin(2\pi t + \pi/4)\). Compare with \(x = A\sin(\omega t + \phi)\): \(A=6\) m, \(\omega=2\pi\) rad/s. \(v_{\max} = A\omega = 12\pi\) m/s.

Step 1: Write the final answer. \(\boxed{A=6\,\text{m},\ v_{\max}=12\pi\,\text{m s}^{-1}}\)