Question

The time period of mass M when displaced from its equilibrium position and then
released for the system as shown in figure is
(a)
2\pi \sqrt{ \frac{m}{k} }2πkm​​
(b)
2\pi \sqrt{ \frac{m}{2k} }2π2km​​
(c)
2\pi \sqrt{ \frac{m}{4k} }2π4km​​
(d)
2\pi \sqrt{ \frac{2m}{k} }2πk2m​​
​

Answer 1

Answer:

When mass M is suspended from the given system as shown in figure, let l be the length through which mass M moves down before it comes to rest. In this situation, both the spring and string be stretched by length l. Since string is inextensible, so spring is stretched by length 2l. The tension along the string and spring is the same.

In equilibrium, Mg=2(k2l)

If mass M is pulled down through small distance x, then

F=Mg−2k(2l+2x)=−4kx .......(i)

F∝x and -ve sign shows that it is directed towards mean position. Hence, the mass executes simple harmonic motion.

For SHM, F=−kx

Comparing (i) and (ii), we get k=4k

∴ Time period, T=2π

4k

M