Question

Orchestra instruments are commonly tuned to match an A-note played by the principal oboe. The Baltimore Symphony Orchestra tunes to an A-note at 440 Hz while the Boston Symphony Orchestra tunes to 442 Hz. If the speed of sound is constant at 343 m/s, find the magnitude of difference between the wavelengths of these two different A-notes. (Enter your answer in m.)

Answer 1

The difference in wavelength between the A-notes at 440 Hz and 442 Hz when the speed of sound is 343 m/s is 3.5 mm.

Step-by-step explanation:

The student is inquiring about the difference in wavelength of two A-notes with different frequencies when the speed of sound is constant. The formula relating the speed of sound (v), frequency (f), and wavelength (λ) is given by v = f λ. For the Baltimore Symphony Orchestra, where A-note = 440 Hz, the wavelength would be λ = v / f = 343 m/s / 440 Hz = 0.7795 m. For the Boston Symphony Orchestra, where A-note = 442 Hz, the wavelength would be λ = v / f = 343 m/s / 442 Hz = 0.7760 m. Thus, the magnitude of difference between the two wavelengths is 0.7795 m - 0.7760 m = 0.0035 m or 3.5 mm.

Answer 2

Δλ = 3*10⁻³ m.

Step-by-step explanation:

• At any wave, there exists a fixed relationship between the speed of the wave, the wavelength, and the frequency, as follows:

`v = \lambda* f (1)`

where v is the speed, λ is the wavelength and f is the frequency.

• Rearranging terms, we can get λ from the other two parameters, as follows:

`\lambda = (v)/(f) (2)`

• Since v is constant for sound at 343 m/s, we can find the different wavelengths at different frequencies, as follows:

`\lambda_(1) =(v)/(f_(1)) = (343m/s)/(440(1/s)) = 0.779 m (3)`

`\lambda_(2) =(v)/(f_(2)) = (343m/s)/(442(1/s)) = 0.776 m (4)`

• The difference between both wavelengths, is just the difference between (3) and (4):

`\Delta \lambda = \lambda_(1) - \lambda_(2) = 0.779 m - 0.776m = 3e-3 m (5)`

⇒ Δλ = 3*10⁻³ m.