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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. 
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A
12πK4M
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B
12π4KM
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C
12πK7M
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D
12π7KM
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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=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'
$$F = -(k+3k)x$$
$$\Rightarrow ma+4kx=0$$
$$\Rightarrow a+\left( \dfrac { 4k }{ m }  \right) x=0\\ \Rightarrow a+{ \omega  }^{ 2 }x=0$$
Here $$\omega =\sqrt { \dfrac { 4k }{ m }  } $$ = angular frequency of oscillation.
frequency $$f=\dfrac { \omega  }{ 2\pi  } =\dfrac { 1 }{ 2\pi  } \sqrt { \dfrac { 4k }{ m }  } $$

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Physics

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