Question

A solid ball of mass $m$ is allowed to fall from a height $h$ to a pan suspended with a spring of spring constant $k$. Assume the ball does not rebound and pan is massless, then amplitude of the oscillation is

• A. $\frac{mg}{k}$
• B. $\frac{mg}{k}{(1+\frac{2hk}{mg})}^{3/2}$
• C. $mg\sqrt{1+\frac{1+2hk}{mg}}$
• D. $\frac{mg}{k}\sqrt{1+\frac{2hk}{mg}}$

Answer 1

The correct option is D $\frac{mg}{k}\sqrt{1+\frac{2hk}{mg}}$

Gravitation on a ball is $mg$, it will exert a force on string that will be equal to $mg$. Spring is extended by ${}^{\prime }{x}^{\prime }$. So,
$kx=mg(k=$ spring constant)
Taking point of hanging as reference.
P.E. $=mg(H+A+x)$
Here $A=$ amplitude of SHM when spring is further disturbed. Now kinetic energy of ball will be,
${\text{(K.E.)}}_{\text{ball}}=\frac{1}{2}k(x+A{)}^2$ (ball will get kinetic energy equal to the P.E. of spring. When it is deformed, P.E. is stored in it)
So, according to conservation of energy K.E. = P.E.
$\Rightarrow \frac{1}{2}k(x+A{)}^2=mg(H+x+A)$
$\Rightarrow \frac{1}{2}k{x}^2+\frac{1}{2}k{A}^2+kxA=mgH+mgx+mgA$
we replace $\left(kx\right)$ by $\left(mg\right)$
So, $\frac{1}{2}k{\left(\frac{mg}{k}\right)}^2+\frac{1}{2}k{A}^2+mgA=mgH+mg\times \frac{mg}{k}+mgA$
$\Rightarrow \frac{1}{2}k{A}^2=mgH+\frac{1}{2}\frac{m^2{g}^2}{k}$
$\Rightarrow k{A}^2=2mgH+\frac{m^2{g}^2}{k}$
$\Rightarrow A^2=\frac{2mgH}{k}+\frac{m^2{g}^2}{k^2}$
$\Rightarrow A=\frac{mg}{k}{(1+\frac{2Hk}{mg})}^{1/2}$