Use the information and table to answer the following question.

Question

A student is using a sound machine to produce sound wave X and sound wave Y.

Sound Wave

Wavelength (m)

Frequency (Hz)

Amplitude (dB)

X

1.5

230

25

Y

2.5

140

50

How could the student manipulate the sounds to make them match in volume but make sound wave X higher pitch than sound wave Y?

Question 13 options:

Sound wave X and sound wave Y should have the same frequency, but sound wave X should have a higher amplitude than sound wave Y.

Sound wave X and sound wave Y should have the same amplitude, but sound wave X should have a higher frequency than sound wave Y.

Sound wave X and sound wave Y should have the same amplitude, but sound wave X should have a higher wavelength than sound wave Y.

Sound wave X and sound wave Y should have the same frequency, but sound wave X should have a lower wavelength than sound wave Y.

Answer 1

To answer this question, we need to understand the relationships between wavelength, frequency, amplitude, and perceived volume and pitch of sound waves.

1. Pitch is determined by frequency: Higher frequency means higher pitch.
2. Volume (loudness) is generally influenced by amplitude: Higher amplitude means louder sound.

From the information given:

• Sound wave X has a frequency of 230 Hz and an amplitude of 25 dB.
• Sound wave Y has a frequency of 140 Hz and an amplitude of 50 dB.

To achieve the goal of having sound wave X at a higher pitch than sound wave Y while matching them in volume, the following conditions must be satisfied:

1. Sound wave X must have a higher frequency than sound wave Y to ensure it is higher in pitch. Since X (230 Hz) is already higher than Y (140 Hz), this condition is satisfied.

2. Sound wave X and sound wave Y should have the same amplitude (loudness) so that they match in volume.

Given this analysis, the correct option is:

Sound wave X and sound wave Y should have the same amplitude, but sound wave X should have a higher frequency than sound wave Y.