Two simple harmonic motions of angular frequencies $100$ and $1000 ext{ rad s}^{-1}$ have the same displacem

Question

Two simple harmonic motions of angular frequencies $100$ and $1000 	ext{ rad s}^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration is:

Accepted Answer

1. **Identify Maximum Acceleration Formula:** In Simple Harmonic Motion, the magnitude of maximum acceleration of a particle is given by $a_{max} = \omega^2 A$, where $\omega$ is the angular frequency and $A$ is the amplitude .
2. **Set up Ratio:** Let the two SHMs have angular frequencies $\omega_1 = 100 	ext{ rad s}^{-1}$ and $\omega_2 = 1000 	ext{ rad s}^{-1}$. They have the same amplitude, so $A_1 = A_2 = A$.
3. **Calculate Ratio:**
   $$ 	ext{Ratio} = \frac{a_{max1}}{a_{max2}} = \frac{\omega_1^2 A_1}{\omega_2^2 A_2} $$
   Since $A_1 = A_2$, this simplifies to:
   $$ 	ext{Ratio} = \left(\frac{\omega_1}{\omega_2}ight)^2 = \left(\frac{100}{1000}ight)^2 $$
   $$ 	ext{Ratio} = \left(\frac{1}{10}ight)^2 = \frac{1}{100} = 1: 10^2 $$

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