Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Three Resonant Frequencies of a String Are 90, 150 and 210 Hz. (A) Find the Highest Possible Fundamental Frequency of Vibration of this String. - Physics

Sum

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?

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Solution

Given:
Let the three resonant frequencies of a string be 

\[f_1  = 90  Hz\] 

\[ f_2  = 150  Hz\] 

\[ f_3  = 210  Hz\]
(a) So, the highest possible fundamental frequency of the string is \[f = 30  Hz\]  because f1f2 and f3 are the integral multiples of 30 Hz.
(b) So, these frequencies can be written as follows:

\[f_1  = 3f\] 

\[ f_2  = 5f\] 

\[ f_3  = 7f\]

Hence, f1f2, and f3 are the third harmonic, the fifth harmonic and the seventh harmonic, respectively.
(c) The frequencies in the string are f, 2f, 3f, 4f, 5f ...
∴ 3f = 2nd overtone and 3rd harmonic
      5f = 4th overtone and 5th harmonic
      7th= 6th overtone and 7th harmonic

(d) Length of the string (L) = 80 cm = 0.8 m
Let the speed of the wave be v.
So, the frequency of the third harmonic is given by:

\[f_1  = \left( \frac{3}{2 \times L} \right)  v\] 

\[ \Rightarrow 90 = \left\{ \frac{3}{\left( 2 \times 80 \right)} \right\} \times v\] 

\[ \Rightarrow v = \frac{\left( 90 \times 2 \times 80 \right)}{3}\] 

\[ = 30 \times 2 \times 80\] 

\[  = 4800  \text{ cm/s }\] 

\[ \Rightarrow v = 48  m/s\]

  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 15 Wave Motion and Waves on a String
Q 46 | Page 326
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