Three resonant frequencies of a string are 90, 150 and 210 Hz. (a) Find the highest possible fundamental frequency of vibration of this string. (b) Which harmonics of the fundamental are the given frequencies? (c) Which overtones are these frequencies? (d) If the length of the string is 80 cm, what would be the speed of a transverse wave on this string?
Answers
(a)The highest possible fundamental frequency of vibration of this string is 30Hz.
(b)Third, fifth and seventh harmonics of the fundamental are the given frequencies.
(c) Second, fourth and sixth overtones are the given frequencies.
(d) the speed of a transverse wave on this string is 48 m/s.
(A)Given, the frequencies are 90,150 and 210 Hz.
The highest possible fundamental frequency would be the HCF of the given frequencies.
90 = 2×3×3×5
150 = 2×3×5×5
210 = 2×3×5×7
So, HCF of 90, 150 and 210 = 2×3×5 = 30
So, highest possible fundamental frequency is 30Hz
So, f = 30Hz
(B)The fundamental frequencies are f',f'' and f'''.
So, f' = 3f [90Hz/3 = 30Hz = f]
f'' =5f [150Hz/5 = 30Hz = f]
f''' = 7f 210Hz/7 = 30Hz = f][
where f', f'' and f''' are the third, fifth and fifth harmonics.
(C)We know the nth overtone is equal to the (n+1) th frequency.
So, 3f = 2nd overtone and 3rd harmonic
5f = 4th overtone and 5th harmonic
7th= 6th overtone and 7th harmonic
(D)Frequency of third harmonic = 3×30Hz = 90Hz
If the speed of transverse wave is v,
then 90Hz = (3/2L)×v
where, L is the length of string = 80cm = 0.8m [1cm = 0.01m]
⇒90 = [3/(2×0.8)]×v
⇒90 = 3v/1.6
⇒ v = (90×1.6)/3 = 48m/s