Question

Problem 6: Frequencies of the string For a string stretched between two supports, two successive standing-wave frequencies are 510 Hz and 680 Hz. There are also other standing waves with higher and lower frequencies. What is the order of what is the length of the string?

Answer 1

Answer 2

The expression for the frequency, "f(n)" of nth harmonic is given by –

f(n) = {(n+1)*V}/(2L) -------------------------------(i)

So –

f(n+1) = {(n+2)*V}/(2L) ----------------------------(ii)

Now do [(ii) – (i)] –

Means –

f(n+1) - f(n) = (V/2L)

We have given that –

f(n+1) = 680 Hz

f(n) = 510 Hz

So –

680 – 510 = (V/2L)

=> V / 2L = 170 Hz

Also given that, V = 85 m/s

So, 85 / 2L = 170 Hz

=> 2L = 85 / 170 = 0.50 m

=> L = 0.50/2 = 0.25 m

Hence, length of the string = 0.25 m

Now, put n = 0,1,2,3…….

Frequency of 1^st harmonic = V/2L = 170 Hz

Frequency of 2^nd harmonic = 2*V/2L = 2*170 = 340 Hz

Frequency of 3rd harmonic = 3*V/2L = 3*170 = 510 Hz

Frequency of 4th harmonic = 4*V/2L = 4*170 = 680 Hz

Therefore, the mentioned harmonics are 3^rd and 4^th harmonics.

Answer 3

Frequency of wave produced in stretched string is: fundamental f_0 = U/2L U rightarrow speed of wave L rightarrow length of string = > nth order f_n = n U/2L we given two successive frequency, f_n = 510H_z & f_n + 1 = 680H_z f_n = n U/2L = > 510H_z = n U/2L equation (I) and f_n + 1 = (n + 1) U/2L = > 680H_z = (n + 1) U/2L equation (II) from eqn (I) & (II) n + 1/n = 680/510 = 4/3 3(n + 1) = 4n n = 3 = > first two successive is of order third second two successive is of order fourth \r\n From eqn (I) Given U = 85 m/s f_n = n U/2L 510 = 3 times 85m/s/2 times L L = 0.25m