A string that is stretched between fixed supports separated by 75.0 cmhas resonant frequencies of 420and 315 Hz , with no intermediate resonant frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

Short Answer

Expert verified
  1. The lowest resonant frequency is 105 Hz
  2. The wave speed is 158 m/s

Step by step solution

01

Given data

Resonant frequencyf1=420Hz

Resonant frequencyf2=325Hz

Distance between two supports is l=75cmor0.75m

02

Understanding the concept of resonant frequency  

Using the concept of resonant frequency and using the equation for the resonance wavelength, we can find the lowest resonant frequency. Using this frequency and wavelength, it is possible to find the velocity of the wave.

Formula:

The standing wave equation kx=nπ.........(1)

The wavenumber,k=2πλ.....(2)

The velocity of a body, v=fλ.......(3)

03

Step 3(a): Calculation of lowest resonant frequency

For a fixed length, using equation (2) in equation (1), the resonant wavelength is given as:

2πλL=nπλ=2Ln....(4)

Then using equation (4) in equation (3), the resonant frequency can be found as:

f=nv2L

Therefore, the nth frequency is given as:

fn=nv2L

And, the (n+1)th term of resonant frequency is given as:

f(n+1)=(n+1)v2L

Therefore,

The frequency difference between two successive pair of the harmonic frequencies is given as:

f=f(n+1)-fn=(n+1)v2L-nv2L=v2L=f1....(5)

Given that the frequencies represent the consecutive frequencies, from the equation (5), we can write the lowest resonant frequency as:

f(n+1)-fn=420-315=105Hz

Therefore, the lowest possible resonant frequency is 105 Hz

04

Step 4(b): Calculation of wave speed

The longest possible wavelength should be equal to the length which is twice the distance between the support. Therefore,

λ1=2L

So, if f1is the lowest possible frequency, using equation (3) we can write the speed of the waves is given as:

v=f1×2L=105×2××0.75=157.5m/s158m/s

Hence, the value of wave speed is 158 m/s

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