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
12πK4M
B
12π4KM
C
12πK7M
D
12π7KM

Solution

## The correct option is B $$\displaystyle \frac{1}{2\pi}\sqrt{\frac{4K}{M}}$$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 $$\rightarrow \dfrac { 2k.2k }{ 2k+2k } =k$$" parallel " $$\rightarrow$$ k+2k=3kNow 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'$$F = -(k+3k)x$$$$\Rightarrow ma+4kx=0$$$$\Rightarrow a+\left( \dfrac { 4k }{ m } \right) x=0\\ \Rightarrow a+{ \omega }^{ 2 }x=0$$Here $$\omega =\sqrt { \dfrac { 4k }{ m } }$$ = angular frequency of oscillation.frequency $$f=\dfrac { \omega }{ 2\pi } =\dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 4k }{ m } }$$Physics

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