An instrument plays a frequency of 266 Hz. Another identical instrument plays a frequency of 400 Hz. How do the wavelength of each of these sound waves compare?

The wavelength of a sound wave is inversely proportional to its frequency. This means that as the frequency of a sound wave increases, its wavelength decreases, and vice versa.

Step 1: Understand the relationship The relationship between the speed of sound (v), its frequency (f), and its wavelength (λ) is given by the equation v = f * λ.

Step 2: Apply the relationship If we have two sound waves with frequencies f1 and f2, and wavelengths λ1 and λ2, then according to the equation, we have v = f1 * λ1 and v = f2 * λ2.

Step 3: Compare the wavelengths Since the speed of sound is constant, we can set the two equations equal to each other to compare the wavelengths: f1 * λ1 = f2 * λ2.

Step 4: Solve for λ2 If we solve for λ2, we get λ2 = (f1/f2) * λ1.

Step 5: Substitute the given values Given that f1 = 266 Hz and f2 = 400 Hz, we can substitute these values into the equation to find that λ2 = (266/400) * λ1.

Step 6: Interpret the result This means that the wavelength of the sound wave with a frequency of 400 Hz is smaller than the wavelength of the sound wave with a frequency of 266 Hz. The exact ratio is 266/400, or approximately 0.665. So, the wavelength of the 400 Hz sound wave is about 66.5% of the wavelength of the 266 Hz sound wave.