Question 30: Oscillations

Question

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

Accepted Answer

Here, A small body of mass $$m$$ attached to one 
end of a vertically hanging spring performs 
SHM.
Angular frequency $$=\omega$$, Amplitude $$=a$$.
Under SHM, velocity $$v=\omega \sqrt{a^{2}-y^{2}}$$ After detaching from spring, net downward acceleration of the block $$=\mathrm{g}$$
$$	herefore$$ Height attained by the block $$=h$$.
$$	herefore \quad h=y+\frac{v^{2}}{2 g}$$ or $$h=y+\frac{\omega^{2}\left(a^{2}-y^{2}ight)}{2 g}$$
For $$h$$ to be maximum, $$\frac{d h}{d y}=0, y=y''$$
$$	herefore \quad \frac{d h}{d y}=1+\frac{\omega^{2}}{2 g}\left(-2 y^{\prime}ight)$$
$$0=\frac{1-2 \omega^{2} y^{\prime}}{2 g}$$
$$\frac{\omega^{2} y^{\prime}}{g}=1$$
$$y^{\prime}=\frac{g}{\omega^{2}}$$
Since, $$w^{2} > g$$ (given)
$$	herefore a > \frac{g}{\omega^{2}} 	herefore a > y', y'$$ from mean position $$ < a$$.
Hence, $$y'=\frac{g}{\omega^{2}}$$

Answer

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

Answer

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

Answer

$$y=\frac{g}{3\omega^{2}}$$

Answer

$$y=\frac{4g}{7\omega^{2}}$$

Oscillations

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