Answer:
The bodies will not come back to their starting point at he same time.
Explanation:
Since they are both pulled back at an angle to the vertical, there is a tangential component of acceleration a = gsinθ
When θ = 4 , a = 9.8sin4 = 0.684 m/s²
When θ = 8 , a = 9.8sin8 = 1.364 m/s²
Using s = ut + 1/2at². Where s is the distance covered and t = time taken, u = initial speed = 0 (assumed since they are both released at the same time)
So s = 0 × t + 1/2at² = 1/2at²
s = 1/2at²
t = √2s/a. Now, since s is the same for both swings, it follows that
t ∝ 1/a. Since their accelerations are different, the bodies will not come back to their starting point at he same time.