A question on spring extension

Asked by Pallavi Chaturvedi | 18th Oct, 2013, 01:10: PM

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
½ky2 = 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 Romal Bhansali | 18th Oct, 2013, 07:29: PM

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