Question

A mass undergoes simple harmonic motion (SHM) with an amplitude of 10 cm. Its maximum acceleration is 5 m/s\(^2\). The angular frequency \(\omega\) of the mass is

• A. 5 rad/s
• B. 7 rad/s
• C. 10 rad/s
• D. 2(2)5 rad/s

Hint:

SHM Relationships. \(a = -\omega^2 x\). Max displacement \(|x_{max| = A\). Max velocity \(v_{max = \omega A\) (at x=0). Max acceleration \(a_{max = \omega^2 A\) (at x=\(\pm\)A). Ensure units are consistent (e.g., meters for A if a is in m/s\(^2\)).

Answer 1

The Correct Option is B

In Simple Harmonic Motion (SHM), the acceleration \(a\) is related to the displacement \(x\) by \(a = -\omega^2 x\), where \(\omega\) is the angular frequency.
The acceleration is maximum when the displacement is maximum (i.
e.
, equal to the amplitude \(A\)).
The magnitude of the maximum acceleration (\(a_{max}\)) is: $$ a_{max} = \omega^2 A $$ We are given: Amplitude \(A = 10\) cm = 0.
10 m.
Maximum acceleration \(a_{max} = 5\) m/s\(^2\).
We need to find \(\omega\).
Rearranging the formula: $$ \omega^2 = \frac{a_{max}}{A} $$ $$ \omega^2 = \frac{5 \, \text{m/s}^2}{0.
10 \, \text{m}} = \frac{5}{0.
1} = 50 \, (\text{rad/s})^2 $$ $$ \omega = \sqrt{50} \, \text{rad/s} $$ Calculating the square root: \( \sqrt{49} = 7 \).
So, \(\sqrt{50}\) is slightly more than 7.
$$ \omega \approx 7.
07 \, \text{rad/s} $$ The closest option is 7 rad/s.