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

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.