Question

A point mass oscillates along x-axis according to \[ x=x_0\sin\left(\omega t-\frac{\pi}{6}\right) \] If the acceleration of the point mass is written as \[ a=A\sin(\omega t+\delta), \] then

• A. \(A=x_0,\ \delta=-\dfrac{\pi}{6}\)
• B. \(A=x_0\omega^2,\ \delta=-\dfrac{\pi}{6}\)
• C. \(A=x_0\omega^2,\ \delta=\dfrac{\pi}{6}\)
• D. \(A=x_0\omega^2,\ \delta=\dfrac{5\pi}{6}\)

Hint:

In SHM, \[ a=-\omega^2x. \] A negative sign in trigonometric form can be absorbed using \[ -\sin\theta=\sin(\theta+\pi). \]

Answer 1

The Correct Option is D

Step 1: Write the relation between acceleration and displacement in SHM.
For simple harmonic motion, \[ a=-\omega^2x \] Given: \[ x=x_0\sin\left(\omega t-\frac{\pi}{6}\right) \] Therefore, \[ a=-\omega^2x_0\sin\left(\omega t-\frac{\pi}{6}\right) \]

Step 2: Express acceleration in sine form.
Using the identity \[ -\sin\theta=\sin(\theta+\pi), \] we get \[ a=x_0\omega^2 \sin\left(\omega t-\frac{\pi}{6}+\pi\right) \] \[ a=x_0\omega^2 \sin\left(\omega t+\frac{5\pi}{6}\right) \]

Step 3: Compare with the required form.
Comparing with \[ a=A\sin(\omega t+\delta), \] we obtain \[ A=x_0\omega^2 \] and \[ \delta=\frac{5\pi}{6} \]

Step 4: Final conclusion.
Hence, \[ \boxed{A=x_0\omega^2,\ \delta=\frac{5\pi}{6}} \]