Page 46 - Q & A STPM 2022 Physics

Physics Semester 3 STPM Chapter 21 Sound Waves
Hence fundamental frequency f 0 = 150 Hz
Resonance frequencies of a tube open at both ends are f 0 , 2f 0 , 3f 0 , 5f 0 …
Difference of frequencies between successive overtones = f 0 = (750 – 450) Hz
= 300 Hz
Then the resonance frequencies would be 300 Hz, 600 Hz, 900 Hz, …
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Hence the tube is not open at both ends.

Section B Structured Questions

Question 11
A 30.0 cm long guitar string with a mass per unit length of 9.00 × 10 kg m is
–1
–3
stretched to a tension of 28.0 N.
Find
(a) the fundamental frequency, and
(b) the frequencies of the next three harmonics.

Answer
(a) Fundamental frequency, f 0 = 1 T
μ
2l
1 28.0
= –3 Hz
2(0.30) 9.00 × 10
= 93.0 Hz
(b) The next three harmonics are of frequencies: 2(93.0 Hz), 3(93.0 Hz) and
4(93.0 Hz)
= 186 Hz, 279 Hz, 372 Hz

Question 10

A length of a string is L and its mass per unit length is μ. The string is
stretched and the tension in the string is T .The string vibrates at its fundamental
frequency, f.
(i) Draw a diagram to show the vibration of the string.
(ii) Write the expression for the speed of transverse wave along the string.
(iii) Deduce the expression for the fundamental frequency f in terms of L, m and
T.
(iv) Find in terms of f, the new fundamental frequency when (I) the tension in
the string is doubled, and (II) when the length of the string is halved.
Semester
3

406

021 STPM Q&A Physics T3.indd 406 17/02/2022 10:24 AM