when tuning a piano, amusician plays a note that should be at 110Hz while at the same time tapping a 110Hz tuning fork and holding it next to the strings. He hears beats at 4Hz.

(A) state and explain what frequencies the piano could be producing
(b) explain how the musician now finishes tuning his piano
(C) Draw a sketch graph to show the resultant sound as a variation in displacement against time. Label the time axis with values.

What about question c ?

a. F1 = 110-4 = 106 Hz. = The difference freq.

F2 = 110+4 = 114 Hz = The sum freq.

b. Adjust strings until the notes cannot be heard.

(A) The musician hears beats at 4Hz, which means there is a difference of 4 cycles per second between the frequency of the note played on the piano and the tuning fork. To determine what frequencies the piano could be producing, we need to consider the relationship between beats and frequency.

Beats occur when two sound waves of slightly different frequencies interfere with each other. The frequency of the beats is equal to the difference between the frequencies of the two sound waves.

In this case, the musician plays a note on the piano that should be at 110Hz, and the tuning fork is vibrating at 110Hz as well. However, since the musician hears beats at 4Hz, it means there is a discrepancy between the actual frequency of the piano string and the desired frequency of 110Hz.

Let's calculate the possible frequencies the piano could be producing:

1. The frequency of the piano string can be higher than 110Hz, resulting in a beat frequency of 4Hz lower than the 110Hz tuning fork frequency. Therefore, the possible piano frequencies can be calculated as follows:
Piano Frequency = 110Hz + 4Hz = 114Hz

2. The frequency of the piano string can be lower than 110Hz, resulting in a beat frequency of 4Hz higher than the 110Hz tuning fork frequency. Therefore, the possible piano frequencies can be calculated as follows:
Piano Frequency = 110Hz - 4Hz = 106Hz

So, based on the beats heard at 4Hz, the piano could be producing frequencies of either 106Hz or 114Hz.

(B) To finish tuning the piano, the musician needs to adjust the tension of the piano string until it matches the desired frequency of 110Hz.

If the piano frequency is 106Hz, the musician needs to increase the tension of the string by tightening it, which will raise the pitch until it matches 110Hz.

If the piano frequency is 114Hz, the musician needs to decrease the tension of the string by loosening it, which will lower the pitch until it matches 110Hz.

By continuously adjusting the tension and checking with the tuning fork, the musician can fine-tune the piano until the desired frequency of 110Hz is achieved.

(C) The graph representing the resultant sound as a variation in displacement against time would depend on the specific characteristics of the beats. However, I can provide a simplified illustration where the cyclical variation in displacement is labeled on the time axis.

```
|
+ |
|
| -
-----------|-----------|-----------|-----------
t0 t1 t2
```

In this graph, the displacement represents the amplitude of the sound wave, with "+" indicating positive displacement (compression) and "-" indicating negative displacement (rarefaction). The time axis is labeled as "t0," "t1," and "t2," representing different points in time. Please note that the actual shape of the graph can vary depending on the specific characteristics of the beats.