When a tuning fork of unknown frequency is sounded simultaneously with a 512 Hz tuning fork, 20 beats are heard in 4.0 seconds. What are the possible frequencies of the unknown tuning fork?
F1 = 512 - 20/4 = 507 Hz.
F2 = 512 + 20/4 = 517 Hz.
The column is at 60 cm in length. a)what is the wavelength of the sound wave produce by fork.
b)what is the frequency of fork if air temperature is 13.6 Centigrade.
To find the possible frequencies of the unknown tuning fork, we can use the formula:
Frequency of Beat = | Frequency of Tuning Fork 2 - Frequency of Tuning Fork 1 |
In this case, the frequency of Tuning Fork 1 is 512 Hz and the number of beats heard is 20 in 4.0 seconds.
So, the Frequency of Beat = 20 beats / 4.0 seconds = 5 Hz
Let x be the frequency of the unknown tuning fork.
Using the formula, we have:
| x - 512 | = 5
Now, we can solve this equation by considering both positive and negative values for x.
1. Positive value for x:
x - 512 = 5
x = 5 + 512
x = 517 Hz
2. Negative value for x:
-(x - 512) = 5
-x + 512 = 5
-x = 5 - 512
-x = -507
x = 507 Hz
Therefore, the two possible frequencies of the unknown tuning fork are 517 Hz and 507 Hz.
To find the possible frequencies of the unknown tuning fork, we can use the concept of beats. Beats occur when two sound waves of slightly different frequencies interfere with each other, resulting in periodic variations in their amplitude.
In this case, the known tuning fork has a frequency of 512 Hz. When it is sounded simultaneously with the unknown tuning fork, we hear 20 beats in 4.0 seconds.
The number of beats (B) per second can be calculated using the formula:
B = |f1 - f2|
where f1 and f2 are the frequencies of the two tuning forks.
We know that the number of beats (B) is 20 and the time (t) is 4.0 seconds. Substituting these values into the formula, we have:
20 = |f1 - 512|
Now, we need to consider two scenarios: one where the frequency of the unknown tuning fork (f1) is greater than 512 Hz, and one where it is less than 512 Hz.
In the first scenario, where f1 is greater than 512 Hz, the formula becomes:
20 = f1 - 512
Solving for f1, we get:
f1 = 532 Hz
In the second scenario, where f1 is less than 512 Hz, the formula becomes:
20 = 512 - f1
Solving for f1, we get:
f1 = 492 Hz
Therefore, the possible frequencies of the unknown tuning fork are 532 Hz and 492 Hz.