Cracking the SAT Physics Subject Test

If frequencies f 1 and f 2 match, then the combined waveform doesn’t waver in

amplitude, and no beats are heard. For example, pianos can be tuned using the
phenomenon of beats. A key is struck, and the corresponding tuning fork is struck; if
the piano string is in tune, there should be no beats as the two sounds interfere. If
beats are heard, then the piano tuner tightens or loosens the string and repeats until
no beats are heard.

Questions 14-15

A piano tuner uses a tuning fork to adjust the key that plays the A

note above middle C (whose frequency should be 440 Hz). The

tuning fork emits a perfect 440 Hz tone. When the tuning fork and the

piano key are struck, beats of frequency 3 Hz are heard.

14. What is the frequency of the piano key?


15. If it’s known that the piano key’s frequency is too high, should
    the piano tuner tighten or loosen the wire inside the piano to tune it?

Here’s How to Crack It

14. Since fbeat = 3 Hz, the tuning fork and the piano string are off by
    3 Hz. Since the fork emits a tone of 440 Hz, the piano string must
    emit a tone of either 437 Hz or 443 Hz. Without more information,
    we can’t decide which one is correct. On the test, only one answer
    is correct per question; pick the one that you see in the answers.


15. If we know that the frequency of the tone emitted by the out-of-
    tune string is too high (that is, it’s 443 Hz), we need to find a way to
    lower the frequency. Remember that the resonant frequencies for a
    stretched string fixed at both ends are given by the equation fn =

nv/2L, and that v =. Since f is too high, v must be too high.

To lower v, we must reduce FT. The piano tuner should loosen the

string and listen for beats again, adjusting the string until the beats

disappear.