Question

an object is oscillating in simple harmonic motion with an amplitude a and an angular frequency ω. what should you do to increase the maximum acceleration of the motion?

○ increase a by 10% and decrease ω by 10%.
○ decrease both a and ω by 10%.
○ decrease a by 10% and increase ω by 10%.
○ decrease a by 10% and leave ω unchanged.
○ leave a unchanged and decrease ω by 10%.

Explanation:

Step1: Recall the formula for maximum acceleration in SHM

The formula for the maximum acceleration \(a_{max}\) in simple - harmonic motion is \(a_{max}=\omega^{2}A\).

Step2: Analyze each option

• Option 1: Increase \(A\) by \(10\%\) and decrease \(\omega\) by \(10\%\)

Let \(A_1 = 1.1A\) and \(\omega_1=0.9\omega\). Then \(a_{max1}=(0.9\omega)^{2}(1.1A)=0.891\omega^{2}A<\omega^{2}A\)

• Option 2: Decrease both \(A\) and \(\omega\) by \(10\%\)

Let \(A_2 = 0.9A\) and \(\omega_2 = 0.9\omega\). Then \(a_{max2}=(0.9\omega)^{2}(0.9A)=0.729\omega^{2}A<\omega^{2}A\)

• Option 3: Decrease \(A\) by \(10\%\) and increase \(\omega\) by \(10\%\)

Let \(A_3 = 0.9A\) and \(\omega_3 = 1.1\omega\). Then \(a_{max3}=(1.1\omega)^{2}(0.9A)=1.089\omega^{2}A\)

• Option 4: Decrease \(A\) by \(10\%\) and leave \(\omega\) unchanged

Let \(A_4 = 0.9A\) and \(\omega_4=\omega\). Then \(a_{max4}=\omega^{2}(0.9A)=0.9\omega^{2}A<\omega^{2}A\)

• Option 5: Leave \(A\) unchanged and decrease \(\omega\) by \(10\%\)

Let \(A_5 = A\) and \(\omega_5 = 0.9\omega\). Then \(a_{max5}=(0.9\omega)^{2}A=0.81\omega^{2}A<\omega^{2}A\)

Answer:

Decrease \(A\) by \(10\%\) and increase \(\omega\) by \(10\%\)