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  1. Problems
  2. /A guitarist tunes a string against an electronic tone of 196 Hz…
EasyPhysicsHigh School (11–12)Waves & Sound › Superposition and Standing WavesWaves & Sound › Sound Waves and HearingNumericalConceptual

A guitarist tunes a string against an electronic tone of 196 Hz…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A guitarist tunes a string against an electronic tone of 196 Hz196\ \text{Hz}196 Hz. Together they give 5 beats each second. The guitarist tightens the string a little, and the beats slow to 3 each second. (a) What was the frequency of the string at first? (b) What is it now? (c) What should the guitarist do next?

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

Hz
Hz
Need a reminder? The key ideaShowHide

Two sounds of frequencies f1f_1f1​ and f2f_2f2​ that are close give beats:

fb=∣f1−f2∣Tb=1fb\displaystyle f_b = |f_1 - f_2| \qquad T_b = \frac{1}{f_b}fb​=∣f1​−f2​∣Tb​=fb​1​

You hear the pitch f1+f22\dfrac{f_1 + f_2}{2}2f1​+f2​​. Wax lowers a fork's frequency, filing raises it, tightening raises a string's.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

(a) At first. The string was 201 Hz201\ \text{Hz}201 Hz or 191 Hz191\ \text{Hz}191 Hz. Tightening raises its frequency, and the beats got slower. So it moved towards 196 Hz196\ \text{Hz}196 Hz from below. It was 191 Hz191\ \text{Hz}191 Hz.

(b) Now. 3 beats each second, still below: 196 Hz−3 Hz=194 Hz196\ \text{Hz} - 3\ \text{Hz} = 194\ \text{Hz}196 Hz−3 Hz=194 Hz.

(c) Next. Tighten a little more, until the beats stop. Then the string is at 196 Hz196\ \text{Hz}196 Hz.

Check. Had the string been 201 Hz201\ \text{Hz}201 Hz, tightening would raise it further, and the beats would get faster.

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