question archive Solution : The frequency of the first wire must be 242242 Hz
Solution : The frequency of the first wire must be 242242 Hz
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Solution : The frequency of the first wire must be 242242 Hz.
Explanation : Solving this problem requires knowing the following concepts,
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Speed of propagation of a wave in a string : When a string of linear mass density (mass per unit length) μμ is held at a tension of TT, waves in the string propagate at a speed given by, v=Tμ−−√ v=Tμ Since the tension (TT) of the wire is held constant the speed of propagation of the wave along the string (vv) is also a constant.
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Standing Waves & : The wavelength and frequencies of the various modes of standing waves (harmonics) are related to the string length (LL) as: λn=2Ln; fn=vλn=nv2L=nL.v2 λn=2Ln; fn=vλn=nv2L=nL.v2
Because vv is a constant, the frequency of a particular mode (fnfn) depends only on the string length (LL). A shorter string produces a higher frequency while a long string produces a lower frequency.
fn(L1)=nL1.v2; fn(L2)=nL2.v2; fn(L1)fn(L2)=L2L1−(1)fn(L1)=nL1.v2; fn(L2)=nL2.v2; fn(L1)fn(L2)=L2L1-(1)
- Formation of Beats : Beats form as a result of the superposition of two waves that differ slightly in their frequencies. The beat frequency is equal to the frequency difference between the interacting waves.
fbeat=?ftest−fref? fbeat=?ftest-fref?
This Problem : Let fofo be the frequency of the reference string, which is what we are interested in finding.
As the length of the test string increases from L1=120L1=120 cm to L2=122L2=122 cm, its frequency decreases from f1f1 to f2f2.
From Equation (1)(1) : f1f2=L2L1=122cm120cm=6160−(2)f1f2=L2L1=122cm120cm=6160-(2)
Since there are two beats produced in both cases, the reference frequency fofo must be 22 Hz lower than f1f1 and 22 Hz higher than f2f2. f1=fo+2; f2=fo−2; f1=fo+2; f2=fo-2;
f1f2=fo+2fo−2−(3) f1f2=fo+2fo-2-(3)
Comparing Equations (2)(2) & (3)(3), fo+2fo−2=6061 fo+2fo-2=6061.
Solving this equation for fofo we get fo=242fo=242 Hz.
