A string of length L vibrates with a fundamental frequency f when under a tension of 100 N. What is the fundamental frequency of the string when its tension is increased to 2500 N?
frequency is proportional to square root of tension
f *sqrt (25) = 5 f
To find the fundamental frequency of the string when its tension is increased to 2500 N, we can use the relationship between tension and frequency for a vibrating string.
The relationship between tension, frequency, and linear density (mass per unit length) of a string is given by the equation:
f = (1/2L) * √(T/μ)
where:
f is the fundamental frequency,
L is the length of the string,
T is the tension, and
μ is the linear density.
In this case, we are given that the length of the string, L, is constant. Therefore, the linear density, μ, remains constant as well.
Since we are only interested in finding the change in frequency when the tension increases from 100 N to 2500 N, we can divide the equation for the fundamental frequency by itself to get:
f2/f1 = √(T2/T1)
where:
f2 is the final fundamental frequency,
f1 is the initial fundamental frequency,
T2 is the final tension, and
T1 is the initial tension.
Plugging in the values:
T1 = 100 N
T2 = 2500 N
f2/f1 = √(2500/100) = √25
Taking the square root of 25 gives us 5, so:
f2/f1 = 5
Therefore, the fundamental frequency of the string when its tension is increased to 2500 N is five times the initial fundamental frequency.