Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. (b) is stretched by the same force F.

**(a)** What is the maximum extension of the spring in the two cases?

**(b)** If the mass in Fig.

(a) and the two masses in Fig.

(b) is released, what is the period of oscillation in each case?

Asked by Abhisek | 1 year ago | 82

**(a)** The maximum extension of the spring in fig (a) is x. The F = kx

Therefore, x =\( \dfrac{F}{k}\)

The force on each mass acts as the reaction force acting on the other mass. The two mass behaves as if it is fixed with respect to the other.

Therefore, x = \( \dfrac{F}{k}\)

here k is the spring constant

**(b)** In figure (a) the restoring force on the mass is F = -kx, here x is the extension of the spring.

For the mass (m) of the block, force is written as

F = ma = m (\( \dfrac{d^2x}{dt^2}\))

Therefore, m (\( \dfrac{d^2x}{dt^2}\)) = -kx

(\( \dfrac{d^2x}{dt^2}\)) = (\( \dfrac{-k}{m}\))x = – ω^{2}x

Here, angular frequency of oscillation, ω = \( \sqrt{\dfrac{k}{m}}\)

Time period of oscillation, T =\( \dfrac{2π}{ω}\)

= 2π (\( \sqrt{\dfrac{m}{k}}\))

In Figure (b), the centre of the system is O and there are two springs.

Each spring is of length \( \dfrac{I}{k}\) and it is attached to two masses

Therefore F = – 2kx

here x is the extension of the spring.

F = ma = m (\( \dfrac{d^2x}{dt^2}\))

m (\( \dfrac{d^2x}{dt^2}\)) = -2kx

(\( \dfrac{d^2x}{dt^2}\)) = (\( \dfrac{-2k}{m}\))x= -ω^{2 }x

ω = (\( \sqrt{\dfrac{2k}{m}}\))

T = 2π (\( \sqrt{\dfrac{m}{2k}}\))

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