A string of an instrument is stretched between two points 25.0 cm apart with a tension of 65.0 N. The mass/length of the string is 4.00x10-3 kg/m. The frequency of the mode of vibration with the lowest frequency (the first harmonic) is
To find the frequency of the first harmonic of the string, we can use the formula:
f = (1/2L) * sqrt(T/μ)
where:
f = frequency of the first harmonic
L = length of the string
T = tension in the string
μ = mass per unit length of the string
Given:
L = 25.0 cm = 0.25 m (converted to meters)
T = 65.0 N
μ = 4.00x10^-3 kg/m
Let's substitute these values into the formula:
f = (1/2 * 0.25) * sqrt(65.0 / 4.00x10^-3)
f = (0.5) * sqrt(16250)
f = (0.5) * 127.55
f ≈ 63.8 Hz
Therefore, the frequency of the mode of vibration with the lowest frequency (the first harmonic) is approximately 63.8 Hz.
To find the frequency of the first harmonic, we can use the formula:
f = (1/2L) × √(T/μ)
where:
f is the frequency (in Hz),
L is the length of the string (in meters),
T is the tension applied to the string (in Newtons), and
μ is the mass per unit length of the string (in kg/m).
In this case, we are given:
L = 25.0 cm = 0.25 m (since 1 meter = 100 centimeters)
T = 65.0 N
μ = 4.00 × 10^-3 kg/m
Substituting these values into the formula, we get:
f = (1/2 × 0.25) × √(65.0 / (4.00 × 10^-3))
Now, let's calculate the frequency using this formula.