Question

A flutist assembles her flute in a room where the speed of sound is 342 m/s. When she plays the note A, it is in perfect tune with a 440 Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is 346 m/s.

-got A: How many beats per second will she hear if she now plays the note A as the tuning fork is sounded? 5.15 Beats/s

-now gotta find B:
How far does she need to extend the "tuning joint" of her flute to be in tune with the tuning fork? (answer is d=_____mm)

Answer 1

The flutist needs to extend the 'tuning joint' of her flute by 464 mm to be in tune with the tuning fork.

Step-by-step explanation:

To calculate how far the flutist needs to extend the 'tuning joint' of her flute to be in tune with the tuning fork, we need to use the equation v = f * λ, where v is the speed of sound, f is the frequency, and λ is the wavelength. First, we need to find the wavelength of the note A played by the tuning fork. Given that the speed of sound is 346 m/s and the frequency is 440 Hz, we can rearrange the equation to solve for λ: λ = v / f = 346 m/s / 440 Hz = 0.786 m. Since we know that the flute produces a musical sound with a speed of 320 m/s and a frequency of 256 Hz, we can rearrange the equation again to solve for the wavelength: λ = v / f = 320 m/s / 256 Hz = 1.25 m. Now, to find the extension needed, we can subtract the original wavelength from the desired wavelength: 0.786 m - 1.25 m = -0.464 m. Since we're looking for a positive distance, we take the absolute value: |-0.464 m| = 0.464 m. Finally, to convert the distance to millimeters, we multiply by 1000: 0.464 m * 1000 = 464 mm. Therefore, the flutist needs to extend the 'tuning joint' of her flute by 464 mm to be in tune with the tuning fork.

Answer 2

5.15348 Beats/s

4.55 mm

Step-by-step explanation:

`v_1` = Velocity of sound = 342 m/s

`v_2` = Velocity of sound = 346 m/s

`f_1` = First frequency = 440 Hz

Frequency is given by

`f_2=(v_2)/(2L_1)\\\Rightarrow f_2=(346)/(2* 0.38863)\\\Rightarrow f_2=445.15348\ Hz`

Beat frequency is given by

`|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s`

Beat frequency is 5.15348 Hz

Wavelength is given by

`\lambda_1=(v_1)/(f)\\\Rightarrow \lambda_1=(342)/(440)\\\Rightarrow \lambda_1=0.77727\ m`

Relation between length of the flute and wavelength is

`\lambda_1=2L_1\\\Rightarrow L_1=(\lambda_1)/(2)\\\Rightarrow L_1=(0.77727)/(2)\\\Rightarrow L_1=0.38863\ m`

At v = 346 m/s

`\lambda_2=(v_2)/(f)\\\Rightarrow \lambda_2=(346)/(440)\\\Rightarrow \lambda_1=0.78636\ m`

`L_2=(\lambda_2)/(2)\\\Rightarrow L_2=(0.78636)/(2)\\\Rightarrow L_2=0.39318\ m`

Difference in length is

`\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm`

It extends to 4.55 mm