Question

A body of mass $5gram$ is executing S.H.M. about a fixed point $O$. With an amplitude of $10cm$, its maximum velocity is $100cm/s$. Its velocity will be $50cm/s$ at a distance (in $cm$):

• A. $5\sqrt{2}$
• B. $5$
• C. $5\sqrt{3}$
• D. $10\sqrt{2}$

Answer 1

Answer: (C)

Let the spring constant of the spring be $K$

Given : velocity at mean position $v_m=100cm/s$

Amplitude $A=10cm$

Applying conservation of energy at mean position and at the extreme position,

$K.E_o+P.E_o=K.E_A+P.E_A$

$\frac{1}{2}m{v}_m^2+0=0+\frac{1}{2}K{A}^2$

$\frac{1}{2}\times 5\times (100{)}^2+0=0+\frac{1}{2}K(10{)}^2⟹K=500dynes/cm$

Velocity of body is $50cm/s$ at point B which lies $x$ cm away from point O.

Now applying conservation of energy at mean position and at point B,

$K.E_o+P.E_o=K.E_B+P.E_B$

$\frac{1}{2}\times 5\times (100{)}^2+0=\frac{1}{2}\times 5\times (50{)}^2+\frac{1}{2}\times 500\times x^2$

$⟹x=5\sqrt{3}cm$