Question

A spring of force constant $\alpha$ has two blocks of same mass M connected to each end of the spring as shown in the figure. Same force f is applied on each end of the spring. if the masses are released, then the period of vibration is:

• A. $2\pi \sqrt{\frac{M}{2\alpha }}$
• B. $2\pi \sqrt{\frac{M}{\alpha }}$
• C. $2\pi \sqrt{\frac{2\alpha M}{{\alpha }^2}}$
• D. $2\pi \sqrt{\frac{2{\alpha }^2}{2\alpha }}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{M}{2\alpha }}$

As forces on two its is same, the net force on string 'zero'.Hence we can say C.M.of spring fixed. With this fixed point, we can treat this problem as two string problem. The strings have length 'l/2' the total length of the string. If $\alpha$ is spring constant of half string then

$\alpha =\frac{{\alpha }^1{\alpha }^1}{{\alpha }^1+{\alpha }^1}={\alpha }^1=2\alpha$

Hence time period of each mass $T=2\pi \sqrt{\frac{M}{\alpha }}\Rightarrow T=2\pi \sqrt{\frac{M}{2\alpha }}$