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A question on spring extension

Asked by Pallavi Chaturvedi 18th October 2013, 1:10 PM
Answered by Expert
Answer:
When body is lowered to mean position, its extension is x.
According to Hooke's law:
 
F = kx = mg
Thus, we have
 
x = mg/k                        ...... (1)
 
Now, when the spring is allowed to fall under gravity, it oscillates about its mean position.
 
Let y be the amplitude of vibration of spring.
 
Then, by conservation of energy at the lowest point
 
ky2 = mgy
 
Thus, we have
 
y = 2mg/k                        ...... (2)
 
Hence, from (1) and (2), we can say that
 
y = 2x
 
Hence, correct option is b.
Answered by Expert 18th October 2013, 7:29 PM
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