Question

Four massless springs whose force constants are 2k, 2k, k and 2k respectively are attached to a mass M kept on a frictionless plane (as shown in the figure). If the mass M is displaced in the horizontal direction, then the frequency of the system.

• A. $\frac{1}{2\pi }\sqrt{\frac{K}{4M}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{4K}{M}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{K}{7M}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{7K}{M}}$

Answer 1

The correct option is B $\frac{1}{2\pi }\sqrt{\frac{4K}{M}}$

Figure (2) is a reduced case of fig(1).In it find out the equivalent spring constant and replace spring by a spring which has spring constant equal to equivalent.

For series combination $\to \frac{2k.2k}{2k+2k}=k$

" parallel " $\to$ k+2k=3k

Now if we displace the mass by a distance $x$,then spring on left side applies a restoring force'$-kx$' and spring on right side applies a restoring force '$-3kx$'.

So, total force on mass 'm'

$F=-(k+3k)x$

$\Rightarrow ma+4kx=0$

$\Rightarrow a+\left(\frac{4k}{m}\right)x=0\Rightarrow a+{\omega }^{2}x=0$

Here $\omega =\sqrt{\frac{4k}{m}}$ = angular frequency of oscillation.

frequency $f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{4k}{m}}$