Question

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. a. What is the order of these two harmonics (first, second, etc.)? b. If the speed of the propagation of transverse waves is 85m/s what is the length of the string?

Answer 1

Answer:

A. The order of two harmonics is 3 and 4

B. The length of string is 4 m.

Explanation:

we know that the frequency of nth standing wave is given in terms of frequency of first wave (f1) as:

$f_{n}$ = n f1

Thus, if we let:

$f_{n}$ = n f1 = 510 Hz

f1 = 510 Hz/n   ____ eqn (1)

then, due to successive waves:

$f_{n+1}$ = (n+1) f1 = 680 Hz

f1 = 680 Hz/(n+1)  ____ eqn (2)

Comparing eqn (1) and eqn (2) we get:

510 Hz/n = 680 Hz/ (n+1)

(n+1)/n = 680/510

1 + 1/n = 4/3

1/n = 4/3 - 1

n = 3

Therefore, the two successive given waves are 3rd Harmonic and 4th Harmonic.

Now, using value of n in eqn (1)

f1 = 510 Hz/3

f1 = 170 Hz

And we know that;

f1 = V/2L

Therefore,

170 Hz = (85 m/s)/2L

L = 4 m