To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as
[tex]f = \frac{v}{\lambda}[/tex]
Then the relation between two different frequencies with same wavelength would be
[tex]\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}[/tex]
[tex]\frac{f'}{f} = \frac{v'}{v}[/tex]
The beat frequency heard when the two strings are sounded simultaneously is
[tex]f_{beat} = f-f'[/tex]
[tex]f_{beat} = f(1-\frac{f'}{f})[/tex]
[tex]f_{beat} = f(1-\frac{v'}{v})[/tex]
We have the velocity of the transverse waves in stretched string as
[tex]v = \sqrt{\frac{T}{\mu}}[/tex]
[tex]v = \sqrt{\frac{200N}{\mu}}[/tex]
And,
[tex]v' = \sqrt{\frac{196N}{\mu}}[/tex]
Therefore the relation between the two is,
[tex]\frac{v'}{v} = \sqrt{\frac{192}{200}}[/tex]
[tex]\frac{v'}{v} = \sqrt{0.96}[/tex]
Finally substituting this value at the frequency beat equation we have
[tex]f_{beat} = 590(1-\sqrt{0-96})[/tex]
[tex]f_{beat} = 11.92Hz[/tex]
Therefore the beats per second are 11.92Hz
The amount of beats heard is mathematically given as
f= 11.92Hz
Question Parameter(s):
Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz
identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz
Generally, the equation for the frequency is mathematically given as
[tex]v = \sqrt{\frac{T}{\mu}}[/tex]
Therefore
[tex]v = \sqrt{\frac{200N}{\mu}}[/tex]
[tex]v' = \sqrt{\frac{196N}{\mu}}[/tex]
Hence
[tex]\frac{v'}{v} = \sqrt{0.96}[/tex]
In conclusion
[tex]fb= 590(1-\sqrt{0-96})[/tex]
f= 11.92Hz
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