The frequencies of two sound waves compared using the scale on the screen
Problem Statement
The frequencies of two sound waves compared using the scale on the screen
Problem Statement
Two different sounds were generated in a laboratory, and the sound waves from each were captured by a microphone and displayed on a computer screen. The screen was marked with vertical grid lines at equal intervals, and the time represented by one grid square was the same for both waves.
When the first sound wave was displayed on the screen, it was found that the length of one cycle of the wave (from one peak to the next) corresponded exactly to N1 grid squares on the screen. Let the frequency of this sound wave be f1. Next, when the second sound wave was displayed on the screen, it was found that the length of one cycle of the wave corresponded exactly to N2 grid squares on the screen. Let the frequency of this sound wave be f2.
The unit of frequency f is hertz (Hz), and the relationship between frequency and period (the time taken for one cycle) T is given by f=T1.
Find the difference between the frequencies of the two sound waves (f2−f1), which is the larger frequency minus the smaller one, and determine the natural number that conforms to the following input format.
Constraints
- Time represented by one unit on the screen: Δt=0.00100 s
- Number of grid squares in one period of the first wave: N1=5
- Number of grid squares in one period of the second wave: N2=2
Input Format
Let A be the numerical value of the difference in frequencies f2−f1. Enter the value of A as a natural number.
Solution
Explanation
Determining the Period from the Scale on the Screen
The time represented by one unit on the scale is Δt. Since the period T1 of the first sound wave corresponds to N1 units on the scale, and the period T2 of the second sound wave corresponds to N2 units, the periods are expressed as follows:
T1=N1Δt T2=N2ΔtCalculating the Frequencies
Since frequency is the reciprocal of the period (f=T1), the frequencies f1 and f2 are given by:
f1=T11=N1Δt1 f2=T21=N2Δt1Substituting Values and Calculating the Difference
Substitute the values from the constraints section (Δt=0.00100, N1=5, N2=2) into the equations.
Frequency f1 of the first sound wave:
f1=5×0.001001=0.005001=200 HzFrequency f2 of the second sound wave:
$f_2 = \frac{1}{2 \times 0.00100} = \frac{1}{0.00200} = 500\text{ }\mathrm{Hz}The difference in frequencies f2−f1 is calculated as follows:
f2−f1=500−200=300 HzTherefore, the numerical value of the difference in frequencies, A, is 300. (This problem allows you to understand, both visually and numerically, that the second sound wave—which has a narrower wave width (fewer grid squares) on the screen—vibrates more times per second and is therefore a higher-pitched sound.)
The natural number to enter is 300.