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Home » Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass attached to its free end. A force applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass at either end. Each end of the spring in Fig. (b) is stretched by the same force F.
Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass \mathbf{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.
CBSE | Class 11 | NCERT | Oscillations | Physics
Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass \mathbf{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?

Solution:

(a) The maximum extension of the spring in fig (a) is x.

F=k x
x=F / k

The response force operating on the other mass is equal to the force acting on each mass. The two masses behave as if they are fixed in relation to each other.

Therefore, x=\mathrm{F} / \mathrm{k}

where,

\mathrm{k} is the spring constant

(b) The restoring force on the mass in figure (a) is F=-k x, where x is the spring extension.

For the mass (\mathrm{m}) of the block, force can be calculated as

F=m a=m\left(d^{2} x / d t^{2}\right)

So, m\left(d^{2} x / d t^{2}\right)=-k x

\left(d^{2} x / d t^{2}\right)=(-k / m) x=-\omega^{2} x

Here, angular frequency of oscillation will be \omega=\sqrt{k} / \sqrt{m}

Time period of oscillation will be T=2 \pi / \omega

=2 \pi(\sqrt{m} / \sqrt{k})

The system’s centre is 0 in Figure (b), and there are two springs. Each spring is I / 2 in length and is linked to two masses.

Therefore F=-2 k x

here x is the extension of the spring.

F=m a=m\left(d^{2} x / d t^{2}\right)

m\left(d^{2} x / d t^{2}\right)=-2 k x

\left(d^{2} x / d t^{2}\right)=(-2 k / m) x=-\omega^{2} x

\omega=\sqrt{(2 k / m)}

\mathrm{T}=2 \pi(\sqrt{\mathrm{m}} / \sqrt{2} \mathrm{k})

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