Question

how could the student manipulate (change) the sounds to make them
match in pitch but make sound wave #1 louder than sound wave #2?
a the student should make the waves have the same amplitude but
sound wave #1 have a higher frequency.
b the student should make the waves have the same frequency but
make sound wave #1 have a higher amplitude.
c the student should make the waves have the same wavelength
but sound wave #1 have a higher frequency.
d the student should make the waves have the same frequency but
have sound wave #2 have a higher wavelength.

Explanation:

Step1: Recall the relationship between amplitude and loudness

Amplitude of a sound wave is related to its loudness. Higher amplitude means louder sound.

Step2: Analyze each option

• Option A: If wave #1 has same amplitude as wave #2, they would have same loudness. So this is incorrect.
• Option B: Amplitude is related to loudness, not pitch. So this is incorrect.
• Option C: Wavelength is related to frequency (since \(v = f\lambda\), for a constant speed \(v\) of sound in a medium). But wavelength is not directly related to loudness. So this is incorrect.
• Option D: Since amplitude of wave #1 (\(50\ cm\)) is greater than that of wave #2 (\(20\ cm\)), wave #1 is louder. And from \(v = f\lambda\) (assuming speed \(v\) of sound in air is constant), \(\lambda=\frac{v}{f}\). Wave #1 has \(\lambda_1 = 1.5\ m\) and \(f_1=200\ Hz\), wave #2 has \(\lambda_2 = 3.0\ m\) and \(f_2 = 100\ Hz\). The student can make wave #2 have same frequency as wave #1 by changing its wavelength (using \(v = f\lambda\), if \(f\) is to be same as \(f_1\), \(\lambda\) can be adjusted as \(\lambda=\frac{v}{f_1}\)).

Answer:

D. The student should make the waves have the same frequency but sound wave #1 have a higher amplitude.