Question

Three resonant frequencies of the string are \[90Hz,{\rm{ }}150Hz{\rm{ }}and{\rm{ }}210Hz\].
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 \[80{\rm{ }}cm\], what would be the speed of a transverse wave on this string?

Answer 1

Hint: Here, to solve this question we have to be well known about the resonant frequencies and how they make overtones and harmonics. These concepts should be very clear and must be able to apply in this question.

Formula used:
1. For harmonics:
\[{f_1} = {f_0}\] - first harmonic
\[{f_2} = 2{f_0}\] - second harmonic
\[{f_3} = 3{f_0}\] - third harmonic
And so on…
2. Similarly for overtones:
Second harmonic is called as first overtone, third harmonic is called as second overtone, and so on.
3. Speed of the wave:
\[{f_1} = \dfrac{v}{{2L}}\]

Complete answer:
a. Let us start to solve this question:
Given data:
Rresonant frequencies are given, we have to calculate fundamental frequency from that: \[90Hz,{\rm{ }}150Hz{\rm{ }}and{\rm{ }}210Hz\]
GCD is calculated as: common divisor for all the three is \[30Hz\]
Therefore, the fundamental frequency, \[{f_0}\] is \[30Hz\].

b. Here, we have to find which of the fundamental the given frequencies are. For that we have to use formula that we are going to use for consideration
\[{f_1} = {f_0}\] - first harmonic
Let us consider,\[{f_1} = 90Hz\], then the multiple of the fundamental is given by:
\[{f_1} = 90Hz = 3 \times 30Hz = 3{f_0}\] , that is third harmonic
Similarly, let us consider,
\[{f_2} = 150Hz = 5 \times 30Hz = 5{f_0}\], that is fifth harmonic
Again,
\[{f_3} = 210Hz = 7 \times 30Hz = 7{f_0}\], that is seventh harmonic
c. For overtones, we know that Second harmonic is called as first overtone, and so on
Therefore, for \[{f_1} = 90Hz\]
\[{f_1} = 90Hz = 3 \times 30Hz = 3{f_0}\]
It is third harmonic that means second overtone.
Similarly, \[{f_2} = 150Hz = 5 \times 30Hz = 5{f_0}\] is the 4th overtone.
\[{f_3} = 210Hz = 7 \times 30Hz = 7{f_0}\] is the 6th overtone.

d. Length of the string is \[80{\rm{ }}cm\], we have to calculate speed of transverse wave.
\[L = 80cm = 80 \times {10^{ - 2}}m\]
\[{f_1} = \dfrac{v}{{2L}}\] is the formula that we are going to use here.
\[v\] be the speed of the wave on the string.
 Now put all the values in above formula, we get
\[{f_1} = \dfrac{v}{{2L}} \Rightarrow 90Hz = \dfrac{v}{{2 \times 80 \times {{10}^{ - 2}}m}}\]
\[ \Rightarrow v = 90Hz \times 2 \times 80 \times {10^{ - 2}}m\]
\[ \Rightarrow v = 14400 \times {10^{ - 2}}m/s\]
\[ \Rightarrow v = 144m/s\]
Thus, the speed of wave on the string is given by \[144m/s\]