Question

Figure 14.30 (a) shows a spring of force constant k clamped rigidly at one end and a mass attached to its free end. A force F applied at the free end stretches the spring. Figure 14.30 (b) shows the same spring with both ends free and attached to a mass m , then clamped rigidly at one end and a mass at either end. Each end of the spring in Fig. 14.30(b) is stretched by the same force F.

What is the maximum extension of the spring in the two cases? 

If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?

Answer 1

For the one block system:
When a force F, is applied to the free end of the spring, an extension l, is produced. For
the maximum extension, it can be written as:
F=kl
Where, k is the spring constant
Hence, the maximum extension produced in the spring, l=F/k
For the two block system:
The displacement (x) produced in this case is:
\(x=\frac{l}{2}\)
Net force, F = +2 kx = 2k \(\frac{l}{2}\)
∴l=\(\frac{F}{k}\)
For the one block system:

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

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

Where, x is the displacement of the block in time t

∴ \(m \frac{d^2x}{dt^2}=-kx\)

It is negative because the direction of elastic force is opposite to the direction of
displacement.
\(\frac{d^2x}{dt^2}=-(\frac{k}{m})x=-ω^2x\)
Where, ω²=\(\frac{k}{m}\)
ω=√\(\frac{k}{m}\)
Where, ω is angular frequency of the oscillation
∴Time period of the oscillation, T=\(\frac{2π}{ω}\)
=\(\frac{2π}{\sqrt\frac{k}{m}}=2π\frac{m}{k}\)
For the two block system:
F=\(m\frac{ d^2x}{dt^2}\)
\(m\frac{d^2x}{dt^2}=-2kx\)
It is negative because the direction of elastic force is opposite to the direction of displacement.
\(\frac{d^2x}{dt^2}=-[\frac{2k}{m}]x=-ω^2x\)
Where,
Angular frequency, \(ω=\sqrt\frac{2k}{m}\)
∴Time period \(T=\frac{2π}{ω}=2π\sqrt\frac{m}{2k}\)