Homework SourseA taut string has a tension of 75 N Its mass is 15 kg and it
A taut string has a tension of 75 N. Its mass is 15 kg and itslength is 12 m. Consider the string to be fixed at both ends.
(a) Find the wavespeed, v.
(b) A tuning fork offrequency 225 Hz is momentarily touched to the string, which beginsto vibrate. Find the wavelength and write the wave function,y(x,t), if the amplitude is 7 mm.
(c) Write the wavefunction for the reflected wave. Add the incident and reflectedwave functions to produce a function for the net wave.
(d) Is a standing waveestablished? Explain. What is the transverse displacement at theend of the string, x=L?
Solution
(a)To send a wave along a string there should be tension inthe string which means that it has been stretched and pulled tautby forces at its two ends.The tension in the string is equalto the common magnitude of those two forces.As a wave travels alongthe string,it displaces elements of the string by causingadditional stretching,with adjacent sections of string pulling oneach other because of the tension.Thus,we can associate the tensionin the string with the stretching (elasticity) of thgestring.
The wave speed is given by
v = C * (/)^(1/2)
C is a dimensionless constant that cannot be determined withdimensional analysis.
here,C = 1
or v = (/)^(1/2)
= 75 N and = (m/l) where m = 15 kg and l = 12m
(b)The transverse velocity u is the rate at which thedisplacement y of the element is changing.In general,thatdisplacement is given by
y(x,t) = ym* sin(kx - wt) -----------(1)
here,ym= 7 mm
we know that
F = k * x
or m * g = k * x
or x = (m * g/k) ----------(2)
the angular frequency of the string is
w = (k/m)^(1/2)
or w^2 = (k/m)
or k = w^2 * m -----------(3)
from (2) and (3) we get
x = (m * g/w^2 * m) = (g/w^2) ------------(4)
from equation (1),the wavenumber of the wave is
k = (2/) -------------(5)
from equations (4) and (5) we get
x =
or = (g/w^2) -------------(6)
here g = 9.8 m/s^2 and w = 2f and f = 225 Hz
the wavelength of the wave is given by equation (6).
the wave function of the wave is obtained from equation (1) bysubstituting ym,k and w in equation (1)
or y(x,t) = (7 * 10^-3) * sin(kx - wt)
the value of k is obatined from equation (5)
(c)Let one wave traveling along a stretched string be givenby
y1(x,t) = ym* sin(kx - wt)
and another,shifted from the first,by
y2(x,t) = ym* sin(kx - wt +)
From the superposition principle,the resultant wave is thealgebraic sum of the two interfering waves and hasdisplacement
y\'(x,t) = y1(x,t) + y2(x,t) =ym* sin(kx - wt) + ym* sin(kx - wt + )-----------(1)
we can write the sum of the sines of two angles and as
sin + sin = 2sin(1/2)( + ) *cos(1/2)( - ) -------------(2)
applying this relation to equation (1) leads to
y\'(x,t) = [2ym* cos(1/2)] * sin(kx - wt +(1/2)) -------------(3)
the resultant wave is also a sinusoidal wave travelling in thedirection of increasing x.
The resultant wave differs from the interfering waves in tworespects:(1)its phase constant is (1/2),and (2)its amplitudeym\' is the magnitude of the quantity in thebrackets.
ym\' = |2ym*cos(1/2)| (amplitude)-------------(4)
(d)the amplitude of the resultant wave is twice the amplitudeof either interfering wave.That is greatest amplitude the resultantwave can have,because the cosine terms in equations (3) and (4) hasits greatest value (unity) when = 0.Interference thatproduces the greatest possible amplitude is called fullyconstaructive interference.
the transverse displacement at the end of the string, x=L canbe obtained by substituting x = L,w = 0 and = 0 inequation (3) therefore we get
y\'(x,t) = [2ym* cos(1/2)] * sin(kx - wt +(1/2))
or y\'(x,t) = [2ym* cos(1/2)] * sin(kL - 0 * t+ (1/2) * 0) = [2ym* cos(1/2)] * sin(kL)
