Question

Show that even as well as odd harmonics are present as overtones in modes of vibration of string.

Answer 1

Let us consider a string which is held both sides by rigid support. if we vibrates the string , longitudinal waves are produced then, both the ends are nodes and there is an antinode exactly midpoint of both nodes.

Let length of string L
wave length of Longitudinal wave is $\lambda$

see figure, length of string , l = 1/2 × wavelength = $\frac{\lambda}{2}$
so, fundamental frequency = v/2l , where v is speed of sound.

first overtones , length of string ,l= $\frac{\lambda}{2}+\frac{\lambda}{2}$
l = $\frac{2\lambda}{2}$
now, frequency of first overtone = 2v/2L
e.g., 1st overtones = 2nd harmonic

again, 2nd overtone , length of string , l = $\frac{\lambda}{2}+\frac{\lambda}{2}+\frac{\lambda}{2}$
l = $\frac{3\lambda}{2}$
frequency of end overtone = 3v/2l
so, 2nd overtone = 3rd harmonic

similarly, nth overtone = (n + 1)th harmonic

hence ,it is clear that even as well as odd harmonics are present as overtones in modes of vibration of string.

Answer 2

Explanation:

Answer

16

4.0

Verified answer

case 1 :- before filling prongs

frequency of tuning fork D ,

n_D=340Hz

running fork C produces 8 beat per second with running fork D ,

so, frequency of tuning fork C ,

= (340 ± 8)Hz

n_C

so, frequency of tunning fork C before filing the prongs = 332Hz or 348 Hz

Case 2 :- after filling prongs

number of beats per second = 4

so, frequency of tunning fork C after filling prongs = (340 ± 4) Hz = 344Hz or 336 Hz

we know, when prongs are filled in tunning fork, frequency of tuning fork increases.

if we choose frequency of tunning fork C 348HZ then after filling prongs Frequency should be more than 348Hz but here given 344 or 336 Hz

hence, frequency of tunning fork C ≠ 348 Hz

ANSWER