Question

A particle undergoing SHM follows the position-time equation given as x = A sin \((\omega t+\frac{\pi}{3})\) . If the SHM motion has a time period of T, then velocity will be maximum at time \(t=\frac{T}{β}\) for first time after t = 0. Value of β is equal to

Answer 1

Correct Answer: 3

Simple Harmonic Motion: 

In simple harmonic motion, the displacement of a particle is described by the equation:

\[ x = A \sin\left(\omega t + \frac{\pi}{3}\right) \]

Here:

• \(A\) is the amplitude of oscillation.
• \(\omega\) is the angular frequency.
• \(\frac{\pi}{3}\) is the phase angle.

 

Velocity of the Particle:

The velocity \(v\) of the particle can be obtained by differentiating the displacement \(x\) with respect to time \(t\):

\[ v = \frac{dx}{dt} = A\omega \cos\left(\omega t + \frac{\pi}{3}\right) \]

For the velocity to reach its maximum value, the cosine term must be equal to \(\pm 1\):

\[ \cos\left(\omega t + \frac{\pi}{3}\right) = \pm 1 \]

Finding the Time \(t\):

For the nearest value of \(t\), set:

\[ \omega t + \frac{\pi}{3} = \pi \]

Solving for \(\omega t\):

\[ \omega t = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

Substitute \(\omega = \frac{2\pi}{T}\) (where \(T\) is the time period):

\[ \frac{2\pi}{T} t = \frac{2\pi}{3} \]

Cancel \(2\pi\):

\[ t = \frac{T}{3} \]

Phase Constant \(\beta\):

The phase constant \(\beta\) can be determined from the relation between time and the phase of the motion. Here, \(\beta = 3\) is the corresponding value based on the equation.

Final Answer:

\(\beta = 3\)