http://www.chegg.com/homework-help/questions-and-answers/given-4-tables-wave-mot

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http://www.chegg.com/homework-help/questions-and-answers/given-4-tables-wave-motions-mass-string-350g-n-frequency-hz-wavelenght-m-s-velocity-m-s-2--q6526391 Using the data from the link above answer the following questions: (1) As the mass hanging on the string increased, how did the velocity of the wave change? Explain why this makes sense using the equations for waves. (2) As the mass on the string increased, how did the frequency change in order to create the same number of loops? Explain why this makes sense using the equations for waves. (3) Within each table, the velocity of the waves should all be roughly equal to each other. Why is this? (4) Use the average velocity to calculate the mass density of the string for each table. Give at least 4 significant figures for mass density. Show calculations below. (5) The six strings of a guitar have different thicknesses, but are under the same tension and have the same length. Will a wave travel faster in the thicker strings or the thinner strings? By what factor should the tension change if you want to triple the fundamental frequency that is created on each string. Be sure to explain your answers to both of these questions.

Explanation / Answer

1)

We can clearly see that as the mass is increased, the velocity of wave also increases

velocity of wave is given by : v = sqrt(T/u)

where T =tension in the string. Now, as the mass is increased, the tension increaased, so, the velocity also increses.

2)

Frequency also increases.

frequency is given by : f = v/lambda

lambda = wavelength

Now, as f is directly proportional to v . So, as mass changes , velocity also increases (as shown in previous part) and thus frequency increases

3)

It is because, velocity depends on the mass of the hanging body (which is proportional to T) and mass per unit length (u) . As in ech table, these values do not change, so the velocity remains same.

4)

Table 1 :

average velocity, Vavg = (23.25+22.61+23.91+22.39)/4 = 23.04 m/s

Now, V = sqrt(T/u) <---- u = mass density

T = mg

So, Vavg = sqrt(mg / u)

So, 23.04 = sqrt(0.35*9.8/u)

So, u = 6.46*10^-3 kg/m

Similarly, for Table 2 :

average velocity, Vavg = (26.67+26.05 + 26.99 + 27.84)/4 = 26.89 m/s

Now, V = sqrt(T/u) <---- u = mass density

T = mg

So, Vavg = sqrt(mg / u)

So, 26.89 = sqrt(0.45*9.8/u)

So, u = 6.1*10^-3 kg/m

for Table 3 :

average velocity, Vavg = (28.2 + 27.88 + 28.22 + 28.47)/4 = 28.19 m/s

Now, V = sqrt(T/u) <---- u = mass density

T = mg

So, Vavg = sqrt(mg / u)

So, 28.19 = sqrt(0.55*9.8/u)

So, u = 6.783*10^-3 kg/m

5)

Here T = same for both

but u = mass/ length = larger for thicker string

So, velocity will be smaller for thicker string as v is inversely proportional to sqrt(u)

So, wave will travel faster in thinner string.

f is directly proportional to sqrt(T)

So, as f is 3 times, so T = 3^2 = 9 times <-----answer

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