The prong of a tuning fork moves back and forth when it is set into vibration. The distance the prong moves between its extreme positions is 2.24 mm. If the frequency of the tuning fork is 440.7 Hz, what are the maximum velocity and the maximum acceleration of the prong? Assume SHM.
To find the maximum velocity and maximum acceleration of the prong, we need to use the formulas for simple harmonic motion (SHM).
The maximum velocity (v_max) of an object in SHM is given by the formula:
v_max = A * ω
where A is the amplitude (the maximum distance from the equilibrium position) and ω (omega) is the angular frequency.
In this case, the amplitude of the prong's motion is given as 2.24 mm, which we need to convert to meters:
A = 2.24 mm = 2.24 * 10^(-3) m
The angular frequency (ω) is related to the frequency (f) by the formula:
ω = 2πf
Given that the frequency of the tuning fork is 440.7 Hz, we can calculate ω:
ω = 2π * 440.7
Now we can substitute the values into the equation for v_max:
v_max = 2.24 * 10^(-3) * 2π * 440.7
Calculating this, we find:
v_max ≈ 3.11 m/s
So, the maximum velocity of the prong is approximately 3.11 m/s.
Next, let's find the maximum acceleration (a_max) of the prong. The maximum acceleration is given by the formula:
a_max = A * ω^2
where A is the amplitude and ω is the angular frequency.
We have already found the values for A and ω, so we can substitute them into the equation:
a_max = 2.24 * 10^(-3) * (2π * 440.7)^2
Calculating this, we find:
a_max ≈ 6.90 m/s^2
So, the maximum acceleration of the prong is approximately 6.90 m/s^2.