1- A guitar string is 90.0 cm long and has a mass of 3.02 g . From the bridge to

Question

1- A guitar string is 90.0 cm long and has a mass of 3.02 g . From the bridge to the support post (=) is 60.0 cm and the string is under a tension of 540 N . What are the frequencies of the fundamental and first two overtones. f1, f2, f3 = ??? Hz

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2- A 5.8 kg ball hangs from a steel wire 1.70 mm in diameter and 7.00 m long. What would be the speed of a wave in the steel wire?

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3- A transverse wave on a cord is given by D(x,t)=0.12sin(3.6x41.0t), where D andx are in m and t is in s.

a- At t=7.4×102s, what is the displacement of the point on the cord where x=0.50m?

D(0.50m,7.4×102s) = ????? m

b-At t=7.4×102s, what is the velocity of the point on the cord where x=0.50m?

Dt(0.50m,7.4×102s) = ?????? m/s

Explanation / Answer

Hi

Well in this case some concepts of waves are in order to solve this problem.

1. In this case we are given the total length of the string (90 cm) and the mass of the string (3.02 g) to calculate the linear density(u) of said string:

u = L/m = 0.9 m/ 3.02*10-3 kg = 298 kg/m

Then they give us the value of the tension over the string (540 N), which can be used with the linear density to find the value of the speed of the wave:

v = ( T/ u )1/2 = [540 N/ (298 kg/m)]1/2 = 1.35 m/s

Finally, we can use the value of the length of string that vibrates (60 cm) to find the value of the fundamental frequency, which can be used to find the first two overtones:

fundamental frequency

f1 = v /2l = (1.35 m/s) / (2*0.6 m) = 1.125 s-1

first overtone

f2 = 2 f1 = 2 (1.125 s-1 ) = 2.25 s-1

second overtone

f3 = 3 f1 = 3 (1.125 s-1 ) = 3.375 s-1

2. In this case, we use the value of the weight of the ball that hangs from the wire to find the tension over the string.

T = Mg ; M would be the mass of the ball (in this case we are considering that the mass of the string is very small compared to the ball) and g is the acceleration due to gravity (assuming a value of 9.8 m/s2).

T = (5.8 kg)*(9.8 m/s2) = 56.84 N

Then we can find the linear density of the wire, if we know the density of the steel (7850 kg/m3) and the dimensions of the wire we can use the following expression:

u = pV/L = pS = p[ d2/4 ]= (7850 kg/m3) [ *(1.7*10-3 m)2 /4 ] = 0.0178 kg/m

Finally, using the tension over the string and the linear density to find the speed of the wave:

v = ( T/ u )1/2 = [56.84 N/ (0.0178 kg/m)]1/2 = 56.5 m/s

3. In this case we simply use the equations given:

a) D = 0.12 sin(3.6*0.5 - 41*7.4*10-2 ) = -0.113 m

b) Dt = -4.92 cos(3.6*0.5 - 41*7.4*10-2) = -1.63 m

I hope it helps.

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