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Question
A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4 cm below the point, where it was held in hand.
What is the amplitude of oscillation?
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Solution
When the support of the hand is removed, the body oscillates about a mean position.
Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.
Loss in PE of the block = `mgx` ......(i)
Where m = mass of the block
The gain in elastic potential energy of the spring = `1/2 kx^2` ......(ii)
As the two are equal, conserving mechanical energy,
We get, `mgx = 1/2kx^2` or `x = (2mg)/k` ......(iii)
Now, the mean position of oscillation will be, when the block is balanced by the spring.
If x' is the extension in that case, then F = + kx'
But `F = mg`
⇒ `mg = + kx^'`
or `x^' = (mg)/k` ......(iv)
Dividing equation (iii) by equation (iv),
`x/x^' = ((2mg)/k)/((mg)/k)` = 2
⇒ `x = 2x^'`
But given x = 4 cm ......(maximum extension from the unstretched position)
∴ `2x^'` = 4
∴ `x^' = 4/2` = 2 cm
But the displacement of mass from the mean position to the position when the spring attains its natural length is equal to the amplitude of the oscillation.
∴ A = x' = 2 cm
Where A = amplitude of the motion.
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