A dog tunes one string of a banjo to play at 376 Hz, while a second string is tu

Question

A dog tunes one string of a banjo to play at 376 Hz, while a second string is tuned to play at 493 Hz. The strings are made of the same material, and thus have the same density. Determine the following.

a.) The frequency of the second harmonic of the first string. (include units with answer)

b.) The ratio of the masses of the two strings (m1/m2) if the strings have the same length and are under the same tension.

c.) The ratio of the diameters of the two strings (d1/d2) if the strings have the same length and are under the same tension.

d.) The ratio of the lengths of the two strings (L1/L2) if the strings have the same mass per unit length and are under the same tension.

e.) The ratio of the tensions of the two strings (T1/T2) if the strings have the same mass and length.

I will rate, please show how to solve so I can understand.

Explanation / Answer

a) f2 = 2*f1 = 2*262 = 524Hz (first overtone)

second overtone 3*f1 = 3*262 = 786Hz

first overtone D f2= 2*f1 = 2*294 = 588Hz

second overtone f3 = 3*294 = 882Hz

b) We have f = v/2L but v =sqrt(T/(m/L))

So f = sqrt(T*L/m)/2L

So f = (sqrt(T*L))/2L/sqrt(m)

So fC = (sqrt(T*L))/2L/sqrt(mC)

and fD = (sqrt(T*L))/2L/sqrt(mD)

So MC/MD = (fD/fC)^2 = (294/262)^2 = 1.26

c) Now rewriting f^2 = T*L/m/(4L^2) = T/(4mL)

So LC/LD = (fD/fC)^2 = 1.26

d) f^2 = T*L/m/(4L^2) = T/(4mL)

So TC/TD = (fC/fD)^2 = (262/294)^2 = 0.794

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