Problem 5.1(a) in 8.03 Fall 2004

Question

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Problem 5.1 is about Piano galore, and part (a) basically states that

suppose you hit $C_5$ and you hear maximum sound at 1 second intervals, what then is the difference in tension between the strings of $C_5$? The tension in each string in the piano is about 250 Newtons.

The Solution to this problem is:

The frequency of the n-th mode of a string is

$$\nu_n=\frac{\omega_n}{2 \pi}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}$$
Differentiating with respect to T gives:
$$\frac{d\nu_n}{dT} = \frac{n}{4L} \sqrt{\frac{1}{T\mu}} = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \frac{1}{2T} = \frac{1}{2T} \nu_n$$
We know that $n = 1, T = 250 N, \nu_{c_5} = 512 Hz$ and $d\nu = 0.5 Hz$. Thus, $dT = (0.5)(500)/512 \approx 0.5N$

However I think that $d\nu= 1 Hz$ instead of $0.5Hz$.
We know that two SHM'2 of equal amplitude will cause beats:

$$x=x_1+x_2=A\cos(\omega_1t)+A\cos(\omega_2t)=2A\cos(\frac{\omega_1-\omega_2}{2}t)\cos(\frac{\omega_1+\omega_2}{2}t)$$
The term $\cos(\frac{\omega_1-\omega_2}{2}t)$ takes care of beats and successive zeros of the modulating disturbance is one half period of the modulating facotr. That is $$T_{beats}=\frac{2\pi}{|\omega_1-\omega_2|}=\frac{2\pi}{\Delta\omega}$$
We also have $$\Delta\nu=\frac{\Delta\omega}{2 \pi}$$
From the above two equation, we have $$\Delta\nu=\frac{\Delta\omega}{2 \pi}=\frac{1}{T_{beats}}=\nu_{beats}$$
The problem states that

you hear maximum sound at 1 second intervals

which means $\Delta\nu=\nu_{beats} = 1 Hz$

In addition, in A.P French's book Vibrations and Waves, page 26-27:

...This has the consequence that the beats frequency-as observed aurally, e.g., from two tunig forks-is simply the difference of their individual frequencies and not half this frequency, as might be suggested by a first glance at Eq.(2-5). Thus, to take a specific case, if two tuning forks side by side are vibrating at 255 and 257 vibrations per second, their combined effect would be that of middle C(256 vibrations per second) passing through a maximun of loudness twice every second.