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A standing wave is set up in a string of a variable length and tension by a vibrator of variable frequency . Both ends of the string are fixed . When the vibrator has a frequency f , in a string of length L and under tension T , n antinodes are set up in the string is doubled , by what factor should the frequency be changed so that the same number of antinodesis produced?(b) If the frequency and lenght are held constant , what tension will producen + 1 antinodes ? ( c) If the frequency is tripled and the length of the string is halved , by what factor should the tension be changed so that twice as many antinodes are produced ? |
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Answer» Solution :In (a) , we except a lower frequency to go with a longer WAVELENGTH . In (b) , lower tension should go with lower wave speed for shorter wavelength at constant frequency . In ( C) , we will just have to divide it out. a. We have ` f_(n) = (n)/( 2L) = sqrt ((T)/(alpha))`(i) Keeping ` n , T and mu ` constant , we can create twoequations . ` f_(n) L = (n)/(2) sqrt((T)/(mu)) and F'_(n) L' = (n)/(2) sqrt((T)/(mu))` Dividing the equations gives `(f_(n))/( f'_(n)) = (L')/(L)` If `L' = 2 L , then f' _(n) = 1//2 f_(n)` THEREFORE , in order to double the length but keep teh same number of ANTINODES , the frequency should be halved . b. From Eq. (i), we can hold `L and f_(n)` constant to get `(n')/(n) = sqrt((T)/(T'))` From this relation , we see that the tension must be decreased to `T' = T ((n)/(n + 1))^(2)` to produce ` n +1` antinodes c. The time , we rearrange Eq. (i) to produce `( 2 f_(n) L)/(n) = sqrt ((T)/(mu)) and ( 2 f'_(n) L')/(n') = sqrt ((T')/(mu))` Then dividings gives `(T')/(T) = (( f_(n))/( f_(n)) xx (n)/(n') xx (L')/(L))^(2) = (( 3 f_(n))/( f_(n)))^(2) xx ((n)/( 2n))^(2) xx ((L//2)/(L))^(2) xx ((L//2)/(L))^(2) = (9)/(16)` |
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