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

Question

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

Accepted Answer

1. **Identify the Formula:** The maximum acceleration ($a_{max}$) of a particle executing simple harmonic motion is given by the formula $a_{max} = \omega^2 A$, where $\omega$ is the angular frequency and $A$ is the displacement amplitude .
2. **Analyze Given Data:**
   - Angular frequency of first SHM, $\omega_1 = 100	ext{ rad s}^{-1}$.
   - Angular frequency of second SHM, $\omega_2 = 1000	ext{ rad s}^{-1}$.
   - The amplitude ($A$) is the same for both motions.
3. **Calculate Ratio:**
   $$ \frac{a_{1}}{a_{2}} = \frac{\omega_1^2 A}{\omega_2^2 A} = \left(\frac{\omega_1}{\omega_2}ight)^2 $$
   $$ \frac{a_{1}}{a_{2}} = \left(\frac{100}{1000}ight)^2 = \left(\frac{1}{10}ight)^2 = \frac{1}{100} $$
   - The ratio is $1:10^2$.

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