Question

A solid ball of mass m is made to fall form a height H on a pan suspended through a spring of spring constant K as shown in figure. if the ball does not rebound and the pan is mass less, then amplitude of oscillation is

• A. $\frac{mg}{K}$
• B. $\frac{mg}{K}{(1+\frac{2HK}{mg})}^{1/2}$
• C. $\frac{mg}{K}+{\left(\frac{2HK}{mg}\right)}^{1/2}$
• D. $\frac{mg}{K}[1+{(1+\frac{2HK}{mg})}^{1/2}]$

Answer 1

The correct option is B $\frac{mg}{K}{(1+\frac{2HK}{mg})}^{1/2}$

Gravitation on a ball is mg, it will exert a force on string='mg'.Spring is extended by '$x$' so, Kx = mg k=spring constant

taking point of hanging as reference

P.E = mg(H+A+X) Here A = amplitude of S.H.M when spring further disturbed.

Kinetic Energy of ball

$=\frac{1}{2}K(x+A{)}^2$ (ball will get K.E is equal to the P.E of spring when it is deformed P.E is stored in it)

So,according to consevation 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 $Kx$ by mg

So, $\frac{1}{2}K{\left(\frac{mg}{K}\right)}^2+\frac{1}{2}K{A}^2+mgA=mgH+mg\times \frac{mg}{K}+mgA\Rightarrow \frac{1}{2}\frac{m^2{g}^2}{K}+\frac{1}{2}K{A}^2=mgH+\frac{m^2{g}^2}{K}\Rightarrow \frac{1}{2}K{A}^2=mgH+\frac{1}{2}\frac{m^2{g}^2}{K}\Rightarrow K{A}^2=2mgH+\frac{m^2{g}^2}{K}\Rightarrow A^2=\frac{2mgH}{K}+\frac{m^2{g}^2}{K^2}\Rightarrow A=\frac{mg}{K}{(1+\frac{2HK}{mg})}^{\frac{1}{2}}$