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A String on a cello vibrates in its first normal mode with a frequency of 220 Hz. The vibrating segment is 70.0 cm long and has a mass of 1.20 g . (a) Find the tension in the string . (b) Determine the frequency of vibration when the string vibrates in three segments . |
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Answer» Solution :The tension should be LESS than ` 500 N (~ 100 1b)` since excessiveforce on the four cello strings would break the neck of the instrument . If a string vibrates in three ANTINODES ( instead of ONE for the fundamental mode) , so the frequency should be three times grreater than the fundamental . From the string's length , we can find the wavelength . We then can use the wavelength with the fundamental frequency to find the wavr speed . Finally , we can find the tension fromthe wave speed and the LINEAR mass density of the string . When the string vibrates in the lowest frequency mode , the length of string forms a standing wave where is ` L = lambda//2` so the fundamental harmonic wavelength is `lambda = 2 L = 2( 0.700 m)` `= 1.40 m` and the speed is ` v = f lambda = ( 220 s^(-1)) ( 1.40 m) = 308 m//s` (a) . From the tension equation ` v = sqrt ((T)/( mu)) = sqrt ((T)/( m//L))` We get ` T = v^(2) m//L` or `T = ((308 m//s)^(2) ( 1.20 xx 10^(-3) kg))/( 0.700 m) = 163 N` (b). For the third harmonic , the tension , linear density and speed are the same , but the string vibrates in three segments . Thus , the wavelength is one - third of the fundamental wavelength . ` lambda_(3) = lambda_(1)//3` From the equation ` v= f lambda` , we find that the frequency three times as high . ` f_(3) = (v)/( lambda_(3)) = 3(v)/( lambda_(1)) = 3 f_(1) = 660 Hz` The tension seems reasonable , and the third harmonic is three times the fundamental frequency as excepted . Related to part (b) , some stringed instrument players use a technique to double the frequency of a note by playing a natural harmonic , in effect cutting a vibrating string in half . When the string is lightly touched at its midpoint to form a node there , the second harmonic is formed , and the resulting note is one octave HIGHER ( twice the original fundamental frequency ).  
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