Question

At a given time \( t \), velocity \( v(t) \) and acceleration \( a(t) \) of a particle undergoing simple harmonic motion are given by
\[ v(t) = -100 \sin \left( 20 t + \frac{\pi}{3} \right), \] \[ a(t) = -2000 \cos \left( 20 t + \frac{\pi}{3} \right). \] Assuming all quantities are in SI units, the amplitude of the oscillation is ............

Hint:

The amplitude of simple harmonic motion can be found using the relationship between velocity, acceleration, and amplitude.

Answer 1

Correct Answer: 5

Step 1: Understanding the equations for SHM.
In simple harmonic motion (SHM), the velocity and acceleration are related to the amplitude \( A \) and angular frequency \( \omega \) as follows: \[ v(t) = A \omega \cos(\omega t + \phi), \] \[ a(t) = - A \omega^2 \sin(\omega t + \phi). \]

Step 2: Comparing with the given equations.
The given velocity equation is: \[ v(t) = -100 \sin \left( 20 t + \frac{\pi}{3} \right). \] This suggests that the amplitude \( A = 100 \), as the coefficient of the sine function represents \( A \omega \), and we can infer \( \omega = 20 \).
The given acceleration equation is: \[ a(t) = -2000 \cos \left( 20 t + \frac{\pi}{3} \right). \] The coefficient of the cosine function represents \( A \omega^2 \), so: \[ A \omega^2 = 2000. \] Substituting \( \omega = 20 \): \[ A \times (20)^2 = 2000, \] \[ A \times 400 = 2000, \] \[ A = \frac{2000}{400} = 5. \]

Step 3: Conclusion.
The amplitude of the oscillation is \( \boxed{5} \, \text{m} \).