If the density of the string is 0.017 g/cm, and its tension is 1260 N, what is the fundamental frequency? Answer in units of Hz.

Question

A string is stretched to a length of 171 cm and both ends are fixed. If the density of the string is 0.017 g/cm, and its tension is 1260 N, what is the fundamental frequency? Answer in units of Hz

Answer 1

For a wave with velocity v, wavelength λ, and frequency f, you can calculate f using the generic formula

f = v/λ -------- 1

First, we must recognize that the fundamental vibration mode of a string fixed at both ends will have nodes at the two fixed ends and a crest at the center. In the fundamental mode, a string supports a half wave-length (see a text book picture).

Therefore, if length of the string is L, then the wavelength should be twice that.

i.e. λ = 2.L -------- 2

To calculate the velocity of the wave (needed to calculate frequency using eq 1), we use the formula

v = √(T/µ) -------- 3

where T is the tension in N,

μ is the linear mass density of the string in g/m

Using eqns 2 and 3 in Eqn 1, we get

f = 1/2L* √(T/µ) -------- 4

You should be able to calculate the answer.

Answer 2

Hi Sam A.,

You can check out the mechanics of stretched spring vibration on https://www.animations.physics.unsw.edu.au/jw/wave_equation_speed.htm

note that the constant k is defined such that k*2L (where 2L is the wavelength) = 2*π

in the standard equation for the wave y = sin(kx-ωt) .

That should give you enough to solve for ω , v , and so on.

--Cheers, -- Mr. d.