Question

One end of a light spring of force constant K is fixed to a block of mass M placed on a horizontal frictionless surface, the other end of the spring being fixed to a wall. The spring – block system is executing simple harmonic motion of amplitude A and frequency v. When the block is passing through the equilibrium position, an object of a mass m is gently placed on the block. As a result, the frequency of the system becomes v' and the amplitude becomes A' . The ratio v'/v will be,

• A. ${\left(\frac{M}{M+m}\right)}^{1/2}$
• B. ${\left(\frac{m}{M+m}\right)}^{1/2}$
• C. $\sqrt{\frac{MA}{m{A}^{\prime }}}$
• D. $\left[\frac{(M+m)A^{\prime }}{mA}\right]$

Answer 1

The correct option is A

${\left(\frac{M}{M+m}\right)}^{1/2}$

The frequency of the system before the object is placed on the block is given by

$v=\frac{1}{2\pi }{\left(\frac{k}{M+m}\right)}^{1/2}$

After the object of mass m is placed on the block, the new frequency of the system becomes

$v^{\prime }=\frac{1}{2\pi }{\left[\frac{k}{M+m}\right]}^{1/2}$

$\therefore \frac{v^{\prime }}{v}={\left(\frac{M}{M+m}\right)}^{1/2}$