A thin 5.4-m metal rod sustanins a longitudinal standing wave with vibration antinodes at each end of the rod. There are no other antinodes. The density and Young's modulus of this metal are, respectively, 5200 kg/m^3 and 7.9x10^10 N/m2. What is the frequency of the rod's vibration?
fdb
To find the frequency of the rod's vibration, we can use the wave equation for a stretched string.
The wave equation for a stretched string is given by:
v = √(F/μ)
Where:
- v is the wave velocity
- F is the tension in the string
- μ is the linear mass density of the string
In this case, the rod sustains a longitudinal standing wave, so we need to consider the longitudinal wave velocity. The longitudinal wave velocity (v) can be calculated using Young's modulus (E) and the density of the material (ρ) using the formula:
v = √(E / ρ)
Given:
- Density (ρ) = 5200 kg/m^3
- Young's modulus (E) = 7.9x10^10 N/m^2
Let's calculate the wave velocity using Young's modulus and the density:
v = √(7.9x10^10 N/m^2 / 5200 kg/m^3)
v ≈ √(1.51923x10^7 m^2/s^2)
v ≈ 3899.1 m/s
Now, to find the frequency (f) of the vibration, we need to consider the length of the rod and the number of antinodes.
The velocity of the wave is given by:
v = λf
Where:
- v is the wave velocity
- λ is the wavelength
- f is the frequency
For a standing wave, the number of antinodes is half the number of wavelengths.
Given that there are antinodes at each end of the rod, the number of antinodes is 2.
Since there are no other antinodes, the number of wavelengths is also 2.
Therefore, the total length of the rod (L) is equal to one wavelength:
L = λ
The frequency can be found by rearranging the formula:
f = v / λ
Since λ = L / 2, the formula becomes:
f = v / (L / 2)
Substituting the values:
f = 3899.1 m/s / (5.4 m / 2)
f = 3899.1 m/s / 2.7 m
f ≈ 1440.7 Hz
Therefore, the frequency of the rod's vibration is approximately 1440.7 Hz.