Question

The human eardrum responds to sound by vibrating. If the eardrum moves in simple harmonic motion at a frequency of 2.93kHz and an amplitude of 0.103 nm (roughly the diameter of a single hydrogen atom), what is its maximum speed of vibration? (Amazingly, the ear can detect vibrations with amplitudes even smaller than this!) μm/s

Answer 1

The maximum speed of vibration for the eardrum moving at a frequency of 2.93kHz and an amplitude of 0.103 nm can be calculated using the formula for maximum speed in simple harmonic motion, which involves the angular frequency and the amplitude of the motion.

Step-by-step explanation:

The student asked about the maximum speed of vibration of a human eardrum which is vibrating in simple harmonic motion (SHM). To find this we use the formula for the maximum speed (v_max) in SHM which is v_max = ωA, where ω is the angular frequency and A is the amplitude. The angular frequency is related to the frequency (f) by the equation ω = 2πf. Given the frequency (f) is 2.93kHz and the amplitude (A) is 0.103nm (which is converted to meters), we first convert the frequency to Hertz by multiplying kHz by 1000, then we calculate ω, and finally we calculate v_max.

The maximum speed of vibration of the eardrum is thus the product of the angular frequency and the amplitude. The solution is:

1. Convert frequency from kHz to Hz: 2.93 kHz × 1000 = 2930 Hz.
2. Calculate angular frequency: ω = 2π × 2930 Hz.
3. Convert amplitude from nm to meters: 0.103 nm = 0.103 × 10^-9 meters.
4. Calculate maximum speed: v_max = ωA.