Question

For a particle executing simple harmonic motion determine the ratio of average acceleration of particle from extreme position to equilibrium position to the maximum acceleration

Answer 1

Answer:

1/2

maximum acceleration at Equilibrium position - -w^2A

Minimum at extreme position

0

Answer 2

The ratio of average acceleration of particle from extreme position to equilibrium position to the maximum acceleration is $\mathbf{\frac{1}{2}}$.

Explanation:

1. When a particle motion is simple harmonic motion, its acceleration is

  Maximum acceleration at extreme position $\mathbf{(a_{max})=A\omega ^{2}}$

  Minimum acceleration at mean position $\mathbf{(a_{min})=0}$

2. Where A is amplitude of oscillation. It is maximum displacement of particle from mean position. Its unit is metre.  

3. Where $\mathbf{\omega }$ is angular frequency of oscillation. its unit is $\mathbf{\frac{rad}{s}}$.

4. So average of acceleration is

   $\mathbf{a_{avg}=\frac{a_{max}+a_{min}}{2}}$

   $\mathbf{a_{avg}=\frac{A\omega ^{2}+0}{2}=\frac{A\omega ^{2}}{2}}$        ...1)

5. Now ratio of average acceleration to maximum acceleration

    $\mathbf{Ratio =\frac{a_{avg}}{a_{max}}}$

    $\mathbf{Ratio =\frac{A\omega ^{2}}{2\times A\omega ^{2}}=\frac{1}{2}}$