Question

A standing wave with 3 antinodes oscillates on a string. The speed of waves or the string is 15 m/s and the length of the string is .5 m. What is the frequency of this standing wave?

Answer 1

`f = (v_(ph)n)/(2L) implies f = 45Hz`

Explanation:

I'm not sure how much depth you've covered standing waves in, but I'm going to do the full derivation for my equations anyway - since I like knowing where things come from rather than memorisation. If it's too much just skip to the end.

`color(red)("DERIVATION START")`

Consider a stretched string clamped at both ends, ie `x = 0 and x = L`. We will get reflections at each end and these will induce standing waves.

Consider the allowable cases using complex exponential notation. NB: we tend to use just the real part by convention, although there is no reason we could not use the imaginary part instead. The simplest being wave of angular frequency `omega` with amplitude "a" travelling to the right and amplitude "b" to the left.

`y(x,t) = ae^(i(kx-omegat)) + be^(i(kx+omegat))`

Behaviour at boundaries is

`y(0,t) = y(L,t) = 0 AA t`

If we apply this at x = 0:

`Re[ae^(-iwt) + be^(iwt)] = 0 AA t`

`implies acos(wt) + bcos(wt) = 0 AA t`

`implies a = -b`

This is good, because we would expect a `pi` phase shift on reflection and that's what we've got.

We now apply the condition at x = L:

`y(L,t) = Re[a(e^(i(kL - wt)) - e^(i(kL+wt))] = 0`

`y(L,t) = Re[ae^(ikL)(e^(-iwt) - e^(iwt))]`

Use of Euler's formula should confirm that

`e^(-iwt) - e^(iwt) = -2isin(wt)`

so `y(L,t) = Re[ae^(ikL)(-2isin(wt))] = 0`

Using Euler's formula on `e^(ikL)` gives:

`y(L,t) = Re[-2aisin(wt)(cos(kL) + isin(kL)] = 0`

`implies 2asin(kL)sin(wt) = 0 AA t`

In order for this statement to be true for all times, must have `sin(kL) = 0`.

`implies kL = npi`

`implies k = (npi)/L`

The phase velocity of the wave is given by `v_(ph) = w/k`

`implies w/(v_(ph)) = (npi)/L implies w = (v_(ph)npi)/L`

`w= 2pif implies f = w/(2pi)`

`f = (v_(ph)n)/(2L)`

`color(red)("DERIVATION END")`

There are 3 anti nodes so this is the 3rd harmonic, ie n = 3.

`f = (15*3)/(2*0.5) = 45Hz`