Question

The phase difference between displacement and velocity of a particle in simple harmonic motion is

• A. \( \pi \, \text{rad} \)
• B. \( \frac{3\pi}{2} \, \text{rad} \)
• C. zero
• D. \( \frac{\pi}{2} \, \text{rad} \)

Hint:

In simple harmonic motion, the velocity is always \( \frac{\pi}{2} \) radians ahead of the displacement in phase. This is because the velocity is the derivative of displacement with respect to time.

Answer 1

The Correct Option is D

To understand the phase difference between displacement and velocity in simple harmonic motion (SHM), we begin by examining their equations. For a particle in SHM, the displacement \(x(t)\) as a function of time \(t\) is given by:

\[x(t) = A \cos(\omega t + \phi)\]

where:

• \(A\) is the amplitude
• \(\omega\) is the angular frequency
• \(\phi\) is the initial phase angle

The velocity \(v(t)\) is the derivative of the displacement with respect to time:

\[v(t) = \frac{dx(t)}{dt} = -A \omega \sin(\omega t + \phi)\]

This can be rewritten as:

\[v(t) = A \omega \cos\left(\omega t + \phi + \frac{\pi}{2}\right)\]

From this expression, we observe that the velocity leads the displacement by a phase difference of \(\frac{\pi}{2}\) radians.

Therefore, the phase difference between displacement and velocity in simple harmonic motion is:

Final Answer: \(\frac{\pi}{2} \, \text{rad}\)

Answer 2

In simple harmonic motion (SHM), a particle's displacement, velocity, and acceleration are interrelated, each having a specific phase relationship with the others. The displacement \( x \) of a particle is typically represented as:

\( x(t) = A\cos(\omega t + \phi) \)

where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

The velocity \( v(t) \) is the time derivative of displacement:

\( v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi) \)

In SHM, the velocity lags the displacement by a phase difference of \( \frac{\pi}{2} \, \text{rad} \). To visualize this:

• Displacement \( x(t) = A\cos(\omega t + \phi) \)
• Velocity \( v(t) = -A\omega\sin(\omega t + \phi) = A\omega\cos(\omega t + \phi + \frac{\pi}{2}) \)

As seen, velocity is ahead of displacement in phase by \( \frac{\pi}{2} \, \text{rad} \), given by the identity for cosine and sine:

• \(\cos(\theta + \frac{\pi}{2}) = -\sin(\theta)\)

Thus, the correct answer for the phase difference between displacement and velocity in SHM is: \( \frac{\pi}{2} \, \text{rad} \)