Question

question 4 (1 point)
if the angular frequency of the motion of a simple harmonic oscillator is doubled, by what factor
does the maximum acceleration of the oscillator change?

a) 2

b) it does not change.

c) 4

d) 1/4

e) 1/2

Explanation:

Step1: Recall the formula for maximum acceleration in simple harmonic motion (SHM)

The formula for the acceleration of a simple harmonic oscillator is \( a = -\omega^{2}x \), where \( \omega \) is the angular frequency and \( x \) is the displacement. The maximum acceleration (\( a_{max} \)) occurs when the displacement \( x \) is equal to the amplitude \( A \) (since \( |x| \leq A \) in SHM). So, \( a_{max} = \omega^{2}A \) (the negative sign indicates direction, and we consider the magnitude for maximum acceleration).

Step2: Analyze the change in angular frequency

Let the initial angular frequency be \( \omega_1 \) and the new angular frequency be \( \omega_2 = 2\omega_1 \) (since it is doubled). The initial maximum acceleration is \( a_{max1} = \omega_1^{2}A \). The new maximum acceleration is \( a_{max2} = \omega_2^{2}A \). Substitute \( \omega_2 = 2\omega_1 \) into the formula for \( a_{max2} \):

\( a_{max2}=(2\omega_1)^{2}A = 4\omega_1^{2}A \)

But \( a_{max1}=\omega_1^{2}A \), so \( a_{max2} = 4a_{max1} \). This means the maximum acceleration changes by a factor of 4.

Answer:

c) 4