One end of a spring of natural length λ and spring constant k is fixed at the ground and the other is fitted with a smooth ring of mass m which is allowed to slide on a horizontal rod fixed at a height λ (figure). Initially, the spring makes an angle of θ with the vertical when the system is released from rest. If the speed of the ring when the spring becomes vertical is (2 λ / 3) 
m/s then find the value of angle θ (in degree):
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(53)
Sol. 
mv 2 = 
kx 2
m 
= 
kx 2
⇒ x = 
= 
– λ
cos θ = 
⇒ = 53º
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