31. Tuning fork 1 (420 Hz) is sounded with tuning fork 2 and 20 beats are count in 10s.
a) What are the possible frequencies of tuning fork 2?
b) Tuning fork 2 is sounded with tuning fork 3 (426 Hz) and 12 beats are detected in 3s. what are the possible frequencies of tuning fork 2?
c) Using both (a) and (b), what is the frequency of turning fork 2? Explain your reasoning.
a) To find the possible frequencies of tuning fork 2, we can use the formula:
f2 = (f1 + Δf) or (f1 - Δf)
Where f2 is the frequency of tuning fork 2, f1 is the frequency of tuning fork 1, and Δf is the difference in frequency between the two forks.
Given that the frequency of tuning fork 1 (f1) is 420 Hz and 20 beats are counted in 10 seconds, we can calculate Δf:
Δf = (beat count / time) * 2
Δf = (20 / 10) * 2 = 4 Hz
Substituting the values into the formula, we have:
f2 = 420 + 4 = 424 Hz (possible frequency)
f2 = 420 - 4 = 416 Hz (possible frequency)
b) To find the possible frequencies of tuning fork 3, we use the same formula:
f3 = (f2 + Δf) or (f2 - Δf)
Given that the frequency of tuning fork 2 (f2) is 424 Hz and 12 beats are detected in 3 seconds, we can calculate Δf:
Δf = (beat count / time) * 2
Δf = (12 / 3) * 2 = 8 Hz
Substituting the values into the formula, we have:
f3 = 424 + 8 = 432 Hz (possible frequency)
f3 = 424 - 8 = 416 Hz (possible frequency)
c) To determine the frequency of tuning fork 2, we need to find the value of f2 that is common to both parts a) and b).
From part a), we found two possible frequencies for tuning fork 2: 424 Hz and 416 Hz.
From part b), we found two possible frequencies for tuning fork 2: 432 Hz and 416 Hz.
Comparing the values, we see that the only frequency that appears in both parts a) and b) is 416 Hz. Therefore, the frequency of tuning fork 2 is 416 Hz.
a) To find the possible frequencies of tuning fork 2, we can use the formula:
Beat frequency = | Frequency of tuning fork 1 - Frequency of tuning fork 2 |
In this case, the beat frequency is given as 20 beats in 10s, which means there are 20/10 = 2 beats per second.
Using the formula, we can rearrange it to find the possible frequencies of tuning fork 2:
| Frequency of tuning fork 1 - Frequency of tuning fork 2 | = 2
Since the frequency of tuning fork 1 is 420 Hz, we can substitute it in the equation:
| 420 Hz - Frequency of tuning fork 2 | = 2
Now, we need to solve for the frequency of tuning fork 2 by considering two possible cases:
1) 420 Hz - Frequency of tuning fork 2 = 2
This gives us Frequency of tuning fork 2 = 420 Hz - 2 = 418 Hz.
2) -(420 Hz - Frequency of tuning fork 2) = 2
Solving this equation, we get Frequency of tuning fork 2 = 420 Hz + 2 = 422 Hz.
Therefore, the possible frequencies of tuning fork 2 are 418 Hz and 422 Hz.
b) Similar to part (a), we can use the formula:
Beat frequency = | Frequency of tuning fork 2 - Frequency of tuning fork 3 |
Given that the beat frequency is 12 beats in 3s, we can calculate the beats per second as 12/3 = 4 beats per second.
Using the formula, we can rearrange it to find the possible frequencies of tuning fork 2:
| Frequency of tuning fork 2 - Frequency of tuning fork 3 | = 4
Substituting the frequency of tuning fork 3 as 426 Hz, we get:
| Frequency of tuning fork 2 - 426 Hz | = 4
Now, we solve for the frequency of tuning fork 2 by considering two possible cases:
1) Frequency of tuning fork 2 - 426 Hz = 4
This gives us Frequency of tuning fork 2 = 426 Hz + 4 = 430 Hz.
2) -(Frequency of tuning fork 2 - 426 Hz) = 4
Solving this equation, we get Frequency of tuning fork 2 = 426 Hz - 4 = 422 Hz.
Therefore, the possible frequencies of tuning fork 2 are 430 Hz and 422 Hz.
c) To determine the accurate frequency of tuning fork 2, we need to consider the information from both parts (a) and (b).
From part (a), we found that the possible frequencies of tuning fork 2 were 418 Hz and 422 Hz.
From part (b), we found that the possible frequencies of tuning fork 2 were 430 Hz and 422 Hz.
Looking at these possible frequencies, we can see that 422 Hz is the frequency that is consistent in both cases.
Therefore, based on the given information, the frequency of tuning fork 2 is 422 Hz.
To solve these questions, we will use the concept of beats in sound waves. When two sound waves of slightly different frequencies are played together, they create a phenomenon known as beats. The number of beats per second is equal to the difference in frequency between the two sound waves.
a) To find the possible frequencies of tuning fork 2 when 20 beats are counted in 10 seconds, we need to use the formula:
Number of beats = difference in frequency × time
Here, the number of beats is 20, and the time is 10 seconds. Let's substitute these values into the formula:
20 = difference in frequency × 10
Rearranging the formula, we get:
Difference in frequency = 20/10 = 2 Hz
Now, we can find the possible frequencies of tuning fork 2 by adding or subtracting this difference from the frequency of tuning fork 1.
Possible frequency of tuning fork 2 = frequency of tuning fork 1 ± difference in frequency
= 420 Hz ± 2 Hz
= 418 Hz or 422 Hz
Therefore, the possible frequencies of tuning fork 2 are 418 Hz or 422 Hz.
b) Now, let's find the possible frequencies of tuning fork 2 when 12 beats are detected in 3 seconds using the same approach.
Number of beats = difference in frequency × time
12 = difference in frequency × 3
Rearranging the formula, we get:
Difference in frequency = 12/3 = 4 Hz
Possible frequency of tuning fork 2 = frequency of tuning fork 3 ± difference in frequency
= 426 Hz ± 4 Hz
= 422 Hz or 430 Hz
Therefore, the possible frequencies of tuning fork 2 are 422 Hz or 430 Hz.
c) To find the frequency of tuning fork 2 using both the results from (a) and (b), we need to look for the frequency that is common to both sets of possible frequencies. In this case, we find that 422 Hz is common to both sets.
Therefore, the frequency of tuning fork 2 is 422 Hz.