Question

Identical cellos are being tested. One is producing a fundamental frequency of 130.9Hz on a string that is 1.25m long and has a mass of 109g. on the second cello the same string is fingered to reduce the length that can vibrate. if the beat frequency produced by these two strings is 4.33Hz, what is the vibrating length of the second string?

Answer 1

The beat frequency between the two cellos is 4.33Hz. The frequency of the second cello can be found by subtracting the beat frequency from the fundamental frequency of the first cello. However, without the speed of the wave, we cannot calculate the exact vibrating length of the second string.

Step-by-step explanation:

The beat frequency is the difference between the frequencies of two interacting waves. In this case, the beat frequency between the two cellos is 4.33Hz.

Given that the fundamental frequency of one cello is 130.9Hz and the beat frequency is 4.33Hz, we can subtract the beat frequency from the fundamental frequency to find the frequency of the second cello:

Frequency of second cello = Fundamental frequency of first cello - Beat frequency = 130.9Hz - 4.33Hz = 126.57Hz

Now we can use the formula for the fundamental frequency of a vibrating string to find the vibrating length of the second string:

Fundamental frequency = Speed of wave / (2 * Length)

Since the strings are identical, the speed of the wave is the same for both. Rearranging the formula, we have:

Length = Speed of wave / (2 * Fundamental frequency)

Plugging in the values, we get:

Length = Speed of wave / (2 * 126.57Hz) = Speed of wave / 253.14Hz

However, we don't have the speed of the wave given in the question, so we cannot calculate the exact vibrating length of the second string without that information.

Answer 2

the vibrating length of the second string
`L_2= 1.209 m`

Step-by-step explanation:

Formula for fundamental frequency is

`f=(1)/(2L)\sqrt{(T)/(\mu) }`

from this equation we can say that

frequency is inversely proportional to the length of string

⇒
`(L_2)/(L_1)=(f_1)/(f_2) }`.........................1

Here
`L_1= 1.25 m`

`f_1= 130.9 Hz`

`f_2-f_1= 4.33Hz`

⇒
`f_2= 4.33+130.9= 315.23 Hz`

now putting values in equation 1 we get

`(L_2)/(1.25)=(130.9)/(135.23) }`.

`L_2= 1.209 m`