Waves - Result Question 43

45. A string is stretched between two fixed points separated by 75.0cm. It is observed to have resonant frequencies of 420Hz and 315Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is :

(a) 205Hz

(b) 10.5Hz

(c) 105Hz

(d) 155Hz

[2015 RS]

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Answer:

Correct Answer: 45. (c)

Solution:

  1. (c) Resonant frequencies, for a string fixed at both ends, will be

fn=nν2L where, n=1,2,3.

The difference between two consecutive resonant frequency is,

Afn=fn+1fn(n+1)v2Lnv2L=v2L

which is also the lowest resonant frequency ( n=1 ).

Thus, for the given string the lowest resonant frequency will be

=420Hz315Hz

=105Hz

The difference between any two successive frequencies will be ’ n '

According to question, n=420315=105Hz So the lowest frequency of the string is 105Hz.

In a stretched string all multiples of frequencies can be obtained i.e., if fundamental frequency is n then higher frequencies will be 2n,3n,4n



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