Question

A particle starts SHM from the mean position. It’s amplitude is a and total energy E. At one instant its kinetic energy is $\frac{3E}{4}$ its displacement at this instant is

• A. y = $\frac{a}{\sqrt{2}}$
• B. y = $\frac{a}{2}$
• C. y = $\frac{a}{\sqrt{\frac{3}{2}}}$
• D. y = a

Answer 1

Answer: (B)

y = $\frac{a}{2}$

Answer 2

y = A sin ωt

E = $\frac{1}{2}m\omega^{2}A^{2}$

K.E. = $\frac{1}{2}m\omega^{2}\left( A^{2} - y^{2} \right)$ = $\frac { 3 } { 4 }$ $\frac{1}{2}m\omega^{2}A^{2}$

⇒ A² − y² = $\frac{3}{4}$A²

$\frac{A^{2}}{4} = y^{2}$ ⇒ y = $\frac{A}{2}$