Question

A small body attached to one end of a vertically hanging spring is performing SHM about it's mean position with angular frequency $\omega$ and amplitude a. If at a height $y^{\ast }$ from the mean position the body gets detached from the spring, calculate the value of $y^{\ast }$ so that the height H attained by the mass is maximum. The body does not interact with the spring during it's subsequent motion after detachment. $(a{\omega }^2>g)$.

• A. $y=\frac{2g}{{\omega }^2}$
• B. $y=\frac{3g}{{\omega }^2}$
• C. $y=\frac{5g}{{\omega }^2}$
• D. $y=\frac{g}{{\omega }^2}$

Answer 1

The correct option is D $y=\frac{g}{{\omega }^2}$

Angular frequency $=\omega$

Amplitude $=a$

Under SHM velocity, $v=\omega \sqrt{a^2-y^2}$

After detaching from spring, net downward acceleration of block $=g$

$\therefore$ Height attained by block, $h=y+\frac{v^2}{2g}$ $=y+\frac{\omega \sqrt{a^2-y^2}}{2g}$

For $h$ to be maximum, $\frac{dh}{dy}=0,y=y\ast$

$\frac{dh}{dy}=1+\frac{{\omega }^2}{2g}(-2y^2)=0$

$\Rightarrow 1-2y^2\frac{{\omega }^2}{2g}=0$

or $y\ast =\frac{g}{{\omega }^2}$

Since, $a{\omega }^2>g$ (given)

$\therefore a>\frac{g}{{\omega }^2}$

$\therefore a>y\ast ;$

$y\ast$ from mean position $<a$

Hence, $y\ast =\frac{g}{{\omega }^2}$