Question

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

Explanation:

Step1: Recall the formula for maximum acceleration in SHM

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

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 \(A\) by \(10\%\) and leave \(\omega\) unchanged

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

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

Let \(A_1 = 0.9A\) and \(\omega_1 = 0.9\omega\). Then \(a_{max1}=(0.9\omega)^{2}(0.9A)=0.729\omega^{2}A<\omega^{2}A\)

• Option 4: Leave \(A\) unchanged and decrease \(\omega\) by \(10\%\)

Let \(A_1 = A\) and \(\omega_1 = 0.9\omega\). Then \(a_{max1}=(0.9\omega)^{2}A = 0.81\omega^{2}A<\omega^{2}A\)

• Option 5: Decrease \(A\) by \(10\%\) and increase \(\omega\) by \(10\%\)

Let \(A_1=0.9A\) and \(\omega_1 = 1.1\omega\). Then \(a_{max1}=(1.1\omega)^{2}(0.9A)=(1.21)(0.9)\omega^{2}A = 1.089\omega^{2}A>\omega^{2}A\)

Answer:

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