Tension is maintained in a string by attaching one end to a wall and by hanging a 2.34 kg object from the other end of the string after it passes over a pulley that is 2.00 m from the wall. The string has a mass per unit length of 3.45 mg/m. What is the fundamental frequency of this string?
Please help. do I use f=1/2L sqrt(tension/wavelenght)? Thank you.
To find the fundamental frequency of this string, you can use the formula:
f = (1/2L) * √(Tension / μ)
where:
- f is the fundamental frequency in Hz
- L is the length of the string in meters
- Tension is the tension in the string in Newtons
- μ is the linear mass density of the string in kg/m
First, let's determine the length of the string L, which is the sum of the distance from the pulley to the wall (2.00 m) and the additional length of the string on the other side of the pulley. Since the string passes over the pulley, the additional length is equal to twice the distance from the pulley to the object. Therefore, the total length of the string L = 2.00 m + 2 * 2.00 m = 6.00 m.
Next, let's calculate the tension in the string. The tension is equal to the weight of the object hanging from the string, which can be calculated as follows:
Tension = mass * acceleration due to gravity
Tension = 2.34 kg * 9.81 m/s²
Tension = 22.9974 N
Now, we need to determine the linear mass density μ of the string. Given that the mass per unit length of the string is 3.45 mg/m, we need to convert it to kg/m:
μ = mass per unit length / 1000
μ = 3.45 mg/m / 1000
μ = 3.45 × 10⁻⁶ kg/m
Now we have all the values we need to calculate the fundamental frequency (f):
f = (1/2L) * √(Tension / μ)
f = (1 / (2 * 6.00 m)) * √(22.9974 N / (3.45 × 10⁻⁶ kg/m))
f ≈ 0.084 Hz
Therefore, the fundamental frequency of the string is approximately 0.084 Hz.