he B3 string on a guitar is 63 em long and has a mass of 1-5 g (a) How.docx

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he B3 string on a guitar is 63 em long and has a mass of 1.5 g (a) How much tension is required to tune the string to 257 Hz? [6 points] b) You pluck the string so that it starts to oscillate with an amplitude of 5.0 mm. Write down the mathematical expression y(x.t) that describes the standing wave. 3 points (c) What is the wavelength, frequency, and velocity of the sound wave emitted by the vibrating string? 4 points What is the highest transverse velocity of the oscillating string and which part of the string will move at this highest velocity? points (d) (e) The string can be made to oscillate at twice the frequency by placing one finger near (but not on) the very center of the string before plucking it. Sketch this standing wave on the string and indicate all nodes and anti-nodes. 4 points) Solution length of the string=L=0.63 m mass=m=1.5*0.001 kg part a: frequency of standing wave on a string=sqrt(tension/mass per unit length)/(2*length) =sqrt(tension/(1.5*0.001/0.63))/(2*0.63) => 257=16.265*sqrt(tension) ==>tension=249.67 N part b: amplitude=5 mm wavelength=2*L=1.26 m wave equation=y(x,t)=amplitude*sin(w*t-k*x) where w=2*pi*frequency=1614.8 rad/s k=2*pi/wavelength =4.9867 m^(-1) wave equation=5*sin(1614.8*t-4.9867*x) mm part c: wavelength=1.26 m frequency=257 Hz velocity of sound=wavelength*frequency =323.82 m/s part d: highest velocity=amplitude*w =0.005*1614.8 =8.074 m/s it will be a position where cos(w*t-k*x) is maximum i.e. sin(w*t-k*x) is minimum hence it will be at a node. .

Formation

k=2*pi/wavelength
=4.9867 m^(-1)
wave equation=5*sin(1614.8*t-4.9867*x) mm
part c:
wavelength=1.26 m
frequency=257 Hz
velocity of sound=wavelength*frequency
=323.82 m/s
part d:
highest velocity=amplitude*w
=0.005*1614.8
=8.074 m/s
it will be a position where cos(w*t-k*x) is maximum
i.e. sin(w*t-k*x) is minimum
hence it will be at a node.

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