A pendulum bob is released from angle θ with the vertical as shown in the figure. If it’s acceleration at maximum amplitude is same as at mean position, find θ

The correct answer is 2tan^-1(1/2).

Let the length of the string of the simple pendulum shown in the image be $L$ and the mass of the suspended bob be $m$. The dotted vertical reference line and the angle $\theta$ represent the initial position from which the bob is released from rest.

1. Acceleration at the Maximum Amplitude (Initial Position):
At the maximum amplitude, the bob is released from rest, so its velocity is zero:
$v=0$
Consequently, the centripetal acceleration is:
$a_c=\frac{v^2}{L}=0$
The only acceleration present is the tangential acceleration ($a_t$) acting along the circular path, which is due to the tangential component of gravity:
$a_t=g\sin \left(\theta \right)$
Therefore, the total magnitude of acceleration at the maximum amplitude ($a_{\text{max}}$) is:
$a_{\text{max}}=g\sin \left(\theta \right)$

2. Acceleration at the Mean Position:
At the mean position (where the string becomes vertical, i.e., angle is 0), the tangential component of gravity is zero, so the tangential acceleration is zero ($a_t=0$).
The acceleration is purely centripetal ($a_c$) due to the velocity of the bob at this lowest point:
$a_{\text{mean}}=a_c=\frac{v^2}{L}$
Using the law of conservation of mechanical energy, the kinetic energy gained at the mean position is equal to the gravitational potential energy lost during the descent:
$\frac{1}{2}m{v}^2=mgL(1-\cos (\theta \left)\right)$
Solving for $v^2$:
$v^2=2gL(1-\cos (\theta \left)\right)$
Substituting this into the centripetal acceleration formula gives:
$a_{\text{mean}}=\frac{2gL(1-\cos (\theta \left)\right)}{L}=2g(1-\cos (\theta \left)\right)$

3. Finding θ by Equating the Accelerations:
We are given that the magnitude of acceleration at the maximum amplitude is equal to the magnitude of acceleration at the mean position:
$a_{\text{max}}=a_{\text{mean}}$
Substitute the derived expressions:
$g\sin \left(\theta \right)=2g(1-\cos (\theta \left)\right)$
Divide both sides by $g$:
$\sin \left(\theta \right)=2(1-\cos (\theta \left)\right)$
Apply the trigonometric half-angle identities:
$\sin \left(\theta \right)=2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)$
and
$1-\cos \left(\theta \right)=2{\sin }^2\left(\frac{\theta }{2}\right)$
Substitute these identities into the equation:
$2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)=2\cdot \left[2{\sin }^2\left(\frac{\theta }{2}\right)\right]$
$2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)=4{\sin }^2\left(\frac{\theta }{2}\right)$
Since $\theta \ne 0$, then $\sin \left(\frac{\theta }{2}\right)\ne 0$. We can divide both sides by $2\sin \left(\frac{\theta }{2}\right)$:
$\cos \left(\frac{\theta }{2}\right)=2\sin \left(\frac{\theta }{2}\right)$
Rearranging terms gives:
$\tan \left(\frac{\theta }{2}\right)=\frac{1}{2}$
Taking the inverse tangent of both sides:
$\frac{\theta }{2}={\tan }^{-1}\left(\frac{1}{2}\right)$
$\theta =2{\tan }^{-1}\left(\frac{1}{2}\right)$