Question

A particle starts executing S.H.M. of amplitude a and total energy E. At the instant its kinetic energy is $\frac{3E}{4}$ , its displacement y is given by

• A. $y=\frac{1}{\sqrt{2}}$
• B. $y=\frac{1}{2}$
• C. $y=\frac{a\sqrt{3}}{2}$
• D. $y=a$

Answer 1

The correct option is

B

$y=\frac{1}{2}$

Given,

Total energy $E=\frac{1}{2}m(\omega a{)}^2$

Kinetic energy $=\frac{1}{2}m{\omega }^2\sqrt{a^2-y^2}$

Potential energy $=\frac{1}{2}m{\omega }^2{y}^2$

Since its kinetic energy $=\frac{3}{4}E,$

Then the potential energy will be equal to Toatl energy - Kinetic energy $=\frac{E}{4}$ at that instant

Now we can write,

$\therefore \frac{\frac{1}{2}m{\omega }^2{y}^2}{\frac{1}{2}m{\omega }^2{a}^2}=\frac{1}{4}$

After solving

$\Rightarrow y^2=\frac{1}{4}{a}^2;y=\frac{a}{2}$

Hence, displacement y is equal to $\frac{a}{2}$