A piano tuner hears one beat every 2.0s when trying to adjust two strings, one of which is sounding 440 Hz. How far off in frequency is the other string?
the difference in freq is 1/2 hz
1beat/2s. = 0.5beats/s. = 0.5cycles/s. = 0.5 Hz.
To determine how far off in frequency the other string is, we need to calculate the frequency difference between the two strings based on the beat frequency.
Since the piano tuner hears one beat every 2.0 seconds, we know that the beat frequency is 1/2.0 beat per second or 0.5 beats per second.
Now, let's find the frequency difference between the two strings. We'll use the formula:
Frequency difference = Beat frequency / Number of beats per cycle
In this case, since we have two strings, the number of beats per cycle is 2 (each string contributes one beat).
Frequency difference = 0.5 beats per second / 2 beats per cycle
Frequency difference = 0.25 Hz
Therefore, the other string is off in frequency by 0.25 Hz compared to the 440 Hz string.
To solve this problem, we need to understand the concept of beats in the context of adjusting the frequency of two strings on a piano.
Beats occur when two waves of slightly different frequencies interfere with each other. In this case, the listener (the piano tuner) hears one beat every 2.0s.
We know that the frequency of one string is 440 Hz. Let's assume that the frequency of the other string is f2 Hz.
The number of beats per second (BPS) can be calculated using the formula:
BPS = | f1 - f2 |
where f1 is the frequency of one string and f2 is the frequency of the other string.
Given that there is one beat every 2.0 seconds, we can express this as:
BPS = 1 beat / 2.0s
Now, we can substitute the values into the formula:
1 beat / 2.0s = | 440 Hz - f2 |
To find the frequency difference, we rearrange the formula:
| 440 Hz - f2 | = 1 beat / 2.0s
Next, we can isolate the absolute value term by removing the absolute value notation and considering two cases, positive and negative:
440 Hz - f2 = 1 beat / 2.0s or -(440 Hz - f2) = 1 beat / 2.0s
Let's solve both cases:
1. Case 1:
440 Hz - f2 = 1 beat / 2.0s
Simplifying, we have:
f2 = 440 Hz - 0.5 beats/s
2. Case 2:
-(440 Hz - f2) = 1 beat / 2.0s
Simplifying and rearranging, we have:
f2 = 440 Hz + 0.5 beats/s
So, the other string can be off in frequency by either 440 Hz - 0.5 beats/s or 440 Hz + 0.5 beats/s, depending on the case.
We cannot determine the exact value without knowing the value of one beat in terms of frequency.