Two simple harmonic motions of angular frequency 100 and 1000 rads^ -1 have the same displacement amplitude. The ratio of their maximum acceleration is
SHM Maximum Acceleration Question with Detailed Explanation
Two simple harmonic motions of angular frequencies \(100\ \text{rad/s}\) and \(1000\ \text{rad/s}\) have the same displacement amplitude. The ratio of their maximum accelerations is:
(a) \( \frac{1}{10^3} \)
(b) \( \frac{1}{10^4} \)
(c) \( 1:10 \)
(d) \( \frac{1}{10^2} \)
Detailed Answer:
The maximum acceleration \(a_{\text{max}}\) in simple harmonic motion (SHM) is given by:
\[
a_{\text{max}} = \omega^2 A,
\]
where \(\omega\) is the angular frequency, and \(A\) is the amplitude of displacement.
For the two SHMs:
- First motion: \(\omega_1 = 100\ \text{rad/s}\)
- Second motion: \(\omega_2 = 1000\ \text{rad/s}\)
- Amplitudes: \(A_1 = A_2 = A\)
The maximum accelerations for the two motions are:
\[
a_{\text{max,1}} = \omega_1^2 A \quad \text{and} \quad a_{\text{max,2}} = \omega_2^2 A.
\]
Taking the ratio of their maximum accelerations:
\[
\frac{a_{\text{max,1}}}{a_{\text{max,2}}} = \frac{\omega_1^2}{\omega_2^2}.
\]
Substituting the values of \(\omega_1\) and \(\omega_2\):
\[
\frac{a_{\text{max,1}}}{a_{\text{max,2}}} = \frac{(100)^2}{(1000)^2} = \frac{10^4}{10^6} = \frac{1}{10^2}.
\]
Therefore, the correct option is:
\[
\boxed{\frac{1}{10^2}}.
\]
This result shows that the maximum acceleration of the first motion is 1/100th of the second motion.
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