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: The standard equation of SHM is \[ x=A\sin(\omega t+\phi), \] where \[ A=\text{Amplitude}, \qquad \omega=\text{Angular frequency}. \] The maximum speed is given by \[ v_{\max}=A\omega. \]

Step 1: Compare the given equation with the standard SHM equation. Given, \[ x=6\sin\left(2\pi t+\frac{\pi}{4}\right). \] Comparing with \[ x=A\sin(\omega t+\phi), \] we get \[ A=6\,\text{m}, \] and \[ \omega=2\pi\,\text{rad s}^{-1}. \]

Step 2: Calculate the maximum speed. \[ v_{\max}=A\omega. \] \[ v_{\max}=6(2\pi). \] \[ v_{\max}=12\pi\,\text{m s}^{-1}. \]

Step 3: Write the final answer. \[ \boxed{A=6\,\text{m}} \] \[ \boxed{v_{\max}=12\pi\,\text{m s}^{-1}} \] \[ \boxed{\text{Answer = (C)}} \]