(a) Explain the term resonance and give two examples (b)(i) Describe, with the aid of...
(a) Explain the term resonance and give two examples
(b)(i) Describe, with the aid of a labelled diagram, an experiment to show how the frequency of the note emitted by a vibrating string depends on the length of the string.
(ii) State two precautions necessary to obtain an accurate result.
(c) A sonometer wire is plucked and it vibrates emitting a fundamental note. State the effect on the frequency of the note if the
(i) tension in the wire were made nine times as large with no change in the length of the wire;
(ii) length of the wire were doubled with no change in the tension.
Explanation
(a) Resonance occurs whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses received from some other body or system which is vibrating with the same frequency. At resonance, the amplitude is maximum. Two examples of resonance are tuning a radio and divers board.
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Procedure: In the diagram above the string is kept taut by using a constant weight
(w) between the two bridges A and B. The length, L between the bridges is adjusted until the wire resonates with the vibrating tuning fork. This happens when a paper rider flies off or when a loud sound is heard. Then, measure length, L of the resonating wire and frequency f. The experiment is repeated for at least four other tuning forks of known frequencies. In each case the values of the frequency, f and the corresponding length, L are recorded. A graph of f is plotted against 1/L or L against 1/f. A straight-line graph passing through origin is obtained showing that frequency is proportional to the reciprocal of the length.
(iii) Precautions:
(1) A smooth pulley should be used
(2) the tuning fork should be struck on a rubber bung
(c) For a fundamental note
f\(_o\) = \(\frac{1}{20}\) = \(\sqrt{\frac{T}{m}}\)
where f\(_o\) = fundamental frequency
L = Length of string
T = Tension m = linear density or mass per unit length of the string
(i) \(f_1 = \frac{1}{20} \sqrt{\frac{f}{m}}\) = 3(\(\frac{1}{20} \sqrt{\frac{1}{20}} \sqrt{\frac{1}{m}})\)
f\(_1 = 3f_o\)
i.e = the fundamental frequency, \(f_o\) is tripled
(ii) \(f_2 = \frac{1}{20} (\frac{1}{20} \sqrt{\frac{T}{m}}) = \frac{1}{2} \times \frac{1}{20} \sqrt{\frac{1}{m}} = f_2 = \frac{1}{2} f_o\)
i.e the fundamental frequency, \(f_o\) is halved