Two wires are stretched between two fixed supports and have the same length. one wire a there is a second-harmonic standing wave whose frequency is 660 hz. however, the same frequency of 660 hz is the third harmonic on wire b. (a) is the fundamental frequency of wire a greater than, less than, or equal to the fundamental frequency of wire b? explain. (b) how is the fundamental frequency related to the length l of the wire and the speed v at which individual waves travel back and forth on the wire? (c) do the individual waves travel on wire a with a greater, smaller, or the same speed as on wire b? give your reasoning.

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, :

So we have:

- On wire A, the second-harmonic has frequency of , so the fundamental frequency is:

- On wire B, the third-harmonic has frequency of , so the fundamental frequency is

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b)

For standing waves on a string, the fundamental frequency is given by the formula:

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

We know that the two wires have same length L. For wire A, , while for wave B, , so we can write the ratio between the speeds of the waves in the two wires:

So, the waves travel faster on wire A.

what are the options.

explanation: