Question

A dog tunes one string of a banjo to play at 362 Hz, while a second string is tuned to play at 463 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)	Format Check	7.86 pts.110% 2% try penalty Bonus hint Hints: 0,0	# tries: 0
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.		7.86 pts.110% 2% try penalty Bonus hint Hints: 0,0	# tries: 0
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.		7.86 pts.110% 2% try penalty Bonus hint Hints: 2,0	# tries: 0
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.		7.86 pts.110% 2% try penalty Bonus hint Hints: 0,0	# tries: 0
E.	The ratio of the tensions of the two strings (T1/T2) if the strings have the same mass and length.		7.86 pts.110% 2% try penalty Bonus hint Hints: 0,0	# tries: 0
A.	The frequency of the second harmonic of the first string. (include units with answer)	Format Check	7.86 pts.110% 2% try penalty Bonus hint Hints: 0,0	# tries: 0
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.		7.86 pts.110% 2% try penalty Bonus hint Hints: 0,0	# tries: 0
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.		7.86 pts.110% 2% try penalty Bonus hint Hints: 2,0	# tries: 0
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.		7.86 pts.110% 2% try penalty Bonus hint Hints: 0,0	# tries: 0
E.	The ratio of the tensions of the two strings (T1/T2) if the strings have the same mass and length.		7.86 pts.110% 2% try penalty Bonus hint Hints: 0,0	# tries: 0

Answer 1

Given: frequency of string 1, f1=362Hz

frequency of string 2, f2=463Hz

A: The second harmonic of the first string :

fundamental frequency/ first harmonic = 362 Hz

second harmonic = 2 * 362 = 724 Hz

B: given: l1 = l2 = l; equal tension, T1=T2

to find m1/m2.

we know, tension of a string, T = 4*m*l*f2

where, m= mass of string

l = lenth of string

f= frequency of string

C:

given: l1 = l2 = l; equal tension, T1=T2

to find d1/d2.

we know, T = 4*m*l*f2

as the material of the strings are the same so their density is also equal.

substituting m value in tension equation,

D: given: T1=T2 and

To find, l1/l2

substituting for  

E: given: m1=m2; l1=l2

to find: T1/T2