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
To determine when the velocity of the particle executing simple harmonic motion (SHM) will be maximum, let's start by considering the given position-time equation: \( x = A \sin(\omega t + \frac{\pi}{3}) \). The velocity \( v \) in SHM is the derivative of position \( x \) with respect to time \( t \). Thus, \( v = \frac{dx}{dt} = A \omega \cos(\omega t + \frac{\pi}{3}) \). For maximum velocity, the cosine function should equal \( \pm 1 \), which occurs when the argument of the cosine is \( n\pi \) (where \( n \) is an integer). Set \( \omega t + \frac{\pi}{3} = 0 \) for the first time maximum: \( \omega t = -\frac{\pi}{3} \). Since \( \cos(0) = 1 \), \( t = -\frac{\pi}{3\omega} \) is not valid as velocity at \( t = 0 \). Consider the next occurrence: \( \omega t + \frac{\pi}{3} = \pi \), \( \omega t = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \). Solving for \( t \): \( t = \frac{2\pi}{3\omega} \). The given period \( T \) is related to \( \omega \) as \( T = \frac{2\pi}{\omega} \). Substituting \( \omega \): \( t = \frac{2\pi}{3} \cdot \frac{T}{2\pi} = \frac{T}{3} \). The maximum velocity occurs at \( t = \frac{T}{β} \), Thus, \( β = 3 \). Verification ensures \( β \) falls within the range [3,3], confirming the solution. The value of β is 3.
