When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amplitude of 7.2 x10-7 m and a maximum speed of 3.6 x10-3 m/s. (a) What is the frequency (in Hz) of the eardrum's vibrations

796.18 Hz

Explanation:

Applying,

Maximum velocity = Amplitude×Angular velocity

Therefore,

V' = A(2πf) Equation 1

Where V' = maximum velocity of the eardrum, A = Amplitude of vibration of the eardrum, f = frequency of the eardrum vibration, π = pie

make f the subject of the equation

f = V'/2πA Equation 2

From the question,

Given: V' = 3.6×10⁻³ m/s, A' = 7.2×10⁻⁷ m,

Constant: 3.14.

Substitute these values into equation 2

f = 3.6×10⁻³/( 7.2×10⁻⁷×2×3.14)

f = 796.18 Hz

found in the nucleus

has mass of one amu

what was the answer?