How is the frequency of a stretched string related to:
Its tension?

Explanation

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 two equations .
` 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)`