Question

If the displacement 'x' of a body in motion in terms of time 't' is given by \(x = A\sin(\omega t + \theta)\), then the minimum time at which the displacement becomes maximum is

• A. \( \frac{1}{\omega}\left[\frac{\pi}{2} - \theta\right] \)
• B. \( \frac{1}{\omega}\left[\frac{2\omega}{\pi} - \theta\right] \)
• C. \( \frac{1}{\omega}\left[\frac{\pi}{2} - 1\right] \)
  % Original seems like 1/omega, not omega
• D. \( \frac{1}{\omega}\left[\theta - \frac{\omega}{\pi^2}\right] \)

Hint:

Displacement \(x = A\sin(\omega t + \theta)\). Maximum displacement \(x_{max} = A\). This occurs when \( \sin(\omega t + \theta) = 1 \). The phase angle \( \phi = \omega t + \theta \) must be \( \frac{\pi}{2} + 2n\pi \) for integer \(n\). Solve for \(t\): \( t = \frac{1}{\omega} \left( \frac{\pi}{2} + 2n\pi - \theta \right) \). To find the minimum positive time, choose the smallest integer \(n\) that makes \(t \ge 0\). Often, \(n=0\) is sufficient if \( \frac{\pi}{2} - \theta \ge 0 \).

Answer 1

The Correct Option is A

The displacement is given by \(x = A\sin(\omega t + \theta)\).
The maximum value of displacement \(x\) is \(A\) (since \( \sin(\cdot) \) has a maximum value of 1).
Displacement \(x\) is maximum when \( \sin(\omega t + \theta) = 1 \).
The general solution for \( \sin \phi = 1 \) is \( \phi = 2n\pi + \frac{\pi}{2} \), where \( n \) is an integer.
So, \( \omega t + \theta = 2n\pi + \frac{\pi}{2} \).
We need to find the time \(t\): \[ \omega t = 2n\pi + \frac{\pi}{2} - \theta \] \[ t = \frac{1}{\omega} \left( 2n\pi + \frac{\pi}{2} - \theta \right) \] We are looking for the minimum positive time \(t\).
We need to choose an integer \(n\) such that \(t>0\) and is minimized.
The term \( \frac{\pi}{2} - \theta \) can be positive, negative or zero.
We need \( 2n\pi + \frac{\pi}{2} - \theta>0 \).
Let \( \phi_0 = \frac{\pi}{2} - \theta \).
We need \( 2n\pi + \phi_0>0 \).
The smallest non-negative value for the argument of sine to be 1 is \( \pi/2 \).
So we need \( \omega t + \theta \) to be the smallest value \( \ge \theta \) (if \(t \ge 0\)) of the form \( 2n\pi + \pi/2 \).
If \( \frac{\pi}{2} - \theta \ge 0 \), we can choose \(n=0\).
Then \( t = \frac{1}{\omega} \left( \frac{\pi}{2} - \theta \right) \).
This is positive if \( \frac{\pi}{2}>\theta \).
If \( \frac{\pi}{2} - \theta<0 \) (i.
e.
, \( \theta>\pi/2 \)), then choosing \(n=0\) gives a negative time, which is not physical for "minimum time".
In this case, we need to choose the smallest \(n\) such that \( 2n\pi + \frac{\pi}{2} - \theta>0 \).
If \( \theta \) is, for example, \( \pi \), then \( \frac{\pi}{2} - \pi = -\frac{\pi}{2} \).
We need \( 2n\pi - \frac{\pi}{2}>0 \).
Smallest \(n=1\) gives \( 2\pi - \pi/2 = 3\pi/2 \).
So \( t = \frac{1}{\omega} \frac{3\pi}{2} \).
The option (1) \( t = \frac{1}{\omega}\left[\frac{\pi}{2} - \theta\right] \) assumes that this value of \(t\) will be the minimum positive time.
This is true if \( \frac{\pi}{2} - \theta \) is the smallest non-negative value such that \( \omega t + \theta = \text{principal value for max sine} \).
This implies \( \frac{\pi}{2} - \theta \ge 0 \) or that \( \theta \le \pi/2 \).
The problem usually implies finding the first occurrence for \(t \ge 0\).
The phase \( \omega t + \theta \) must be \( \pi/2, 5\pi/2, 9\pi/2, \dots \) or \( -3\pi/2, -7\pi/2, \dots \) for \( \sin = 1 \).
We need \( \omega t + \theta = \Phi_{max} \), where \( \Phi_{max} \) is the smallest value of \( 2k\pi + \pi/2 \) such that \( t \ge 0 \).
So \( t = \frac{\Phi_{max} - \theta}{\omega} \ge 0 \implies \Phi_{max} \ge \theta \).
We choose \( \Phi_{max} \) to be the smallest angle of the form \( (2n+1/2)\pi \) that is \( \ge \theta \).
If \( \theta \le \pi/2 \), then \( \Phi_{max} = \pi/2 \) is a possible choice (for \(n=0\)).
This yields \( t = (\pi/2 - \theta)/\omega \).
If this \(t<0\), i.
e.
\( \theta>\pi/2 \), we choose the next value, \( \Phi_{max} = 2\pi + \pi/2 = 5\pi/2 \).
Then \(t = (5\pi/2 - \theta)/\omega \).
Given the options, it is implied that \( \frac{\pi}{2} - \theta \ge 0 \).
Or, more generally, \( \frac{1}{\omega} (\frac{\pi}{2}-\theta) \) might be negative, but the question asks for "minimum time", usually implying \(t \ge 0\).
The most straightforward interpretation matching option (1) is that \( (2n\pi + \pi/2 - \theta) \) is minimized for \(n=0\) and is non-negative.