AnalyzingWaves

AnalyzingWaves - previous index next Analyzing Waves on a...

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previous index next Analyzing Waves on a String Michael Fowler 5/30/08 From Newton’s Laws to the Wave Equation Everything there is to know about waves on a uniform string can be found by applying Newton’s Second Law, , to one tiny bit of the string. Well, at least this is true of the small amplitude waves we shall be studying—we’ll be assuming the deviation of the string from its rest position is small compared with the wavelength of the waves being studied. This makes the math simpler, and is an excellent approximation for musical instruments, etc. Having said that, we’ll draw diagrams, like the one below, with rather large amplitude waves, to show more clearly what’s going on. Fm a = G G x y 0 T T x x + Δ x Dynamics of a short segment of string: neglecting gravity, the only forces are the tension forces T acting on the ends. Let’s write down for the small length of string between x and x + Δ x in the diagram above. a = G G Taking the string to have mass density μ kg/m, we have . mx = Δ The forces on the bit of string (neglecting the tiny force of gravity, air resistance, etc.) are the tensions T at the two ends. The tension will be uniform in magnitude along the string, but the string curves if it’s waving, so the two T G vectors at opposite ends of the bit of string do not quite cancel, this is the net force we’re looking for. F G Bearing in mind that we’re only interested here in small amplitude waves, we can see from the diagram (squashing it mentally in the y -direction) that both T G vectors will be close to horizontal, and, since they’re pointing in opposite directions, their sum—the net force —will be very close to vertical: F G
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2 The two almost-parallel T (string tension) vectors add to give a resultant almost-vertical force on the string segment. The vertical component of the tension T G at the x + Δ x end of the bit of string is sin T θ , where is the angle of slope of the string at that end. This slope is of course just , or, more precisely, () / dy x x dx /t a n dy dx = . sin T However, if the wave amplitude is small, as we’re assuming, then is small, and we can take tan sin θθ == , and therefore take the vertical component of the tension force on the string to be . So the total vertical force from the tensions at the two ends becomes / TT d y xx d =+ Δ x ( ) ( ) ( ) 2 2 dy x x dy x d y x FT T x dx dx dx ⎛⎞ =− ⎜⎟ ⎝⎠ G Δ the equality becoming exact in the limit 0 x Δ → . At this point, it is necessary to make clear that y is a function of t as well as of x : ( ) , yy x t = . In this case, the standard convention for denoting differentiation with respect to one variable while the other is held constant (which is the case here—we’re looking at the sum of forces at one instant of time) is to replace / with / dd x x ∂ ∂ .
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This note was uploaded on 12/07/2011 for the course PHYSICS 152 taught by Professor Michaelfowler during the Fall '07 term at UVA.

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AnalyzingWaves - previous index next Analyzing Waves on a...

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