Question

The energy of a particle executing simple harmonic motion is given by E = $A{x}^2+B{v}^2$ where $x$ is the displacement from mean position $x=0$ and $v$ is the velocity of the particle at $x$ then choose the correct statement(s).

• A. amplitude of SHM is $\sqrt{\frac{2E}{A}}$
• B. maximum velocity of the particle during SHM is $\sqrt{\frac{E}{B}}$
• C. time period of motion is $2\pi \sqrt{\frac{B}{A}}$
• D. displacement of the particle is directly proportional to the velocity of the particle.

Answer 1

Answer: (A)

amplitude of SHM is $\sqrt{\frac{2E}{A}}$

The total energy of the particle is given as $E=A{x}^2+B{v}^2$

As the velocity of the particle is maximum when displace is zero i.e $x=0$

Thus $E=A(0{)}^2+B(v_{max}{)}^2⟹v_{max}=\sqrt{\frac{E}{B}}$

When the particle is at extreme position, at that instant its velocity is zero.

Let $R$ be the amplitude of the $SHM$.

Thus $E=A(R{)}^2+B(0{)}^2⟹$ Amplitude of $SHM$, $R=\sqrt{\frac{E}{A}}$

Comparing the energy equation with $E=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2⟹k=2A$ and $m=2B$

Now time period of the oscillation $T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{B}{A}}$