(a) State one condition each necessary for the characteristics each of the following occurrences:
(i) Constructive interference of waves.
(ii) Total internal reflection.
(iii) Production of beats.
(b) In a resonance tube experiment using a tuning fork of frequency 256 Hz, the first position of resonance was 35 cm, the next position was 100 cm. Calculate the velocity of sound in air from the experiment.
(c)(i) State the three classifications of musical instalments
(ii) Give one example each of the classifications stated in (c)(i).
(d) Calculate the critical angle for light traveling from glass to air. [refractive index of glass = 1.5].
(e) The speed of sound in a medium at a temperature of 102 °C is 240 m s\(^{-1}\). If the speed of sound in the medium is 3 10 m s\(^{-1}\). Calculate its temperature
Explanation
(a) Conditions necessary for:
Constructive interference of waves
i. lt must involve at least two waves.
ii. Waves must vibrate in phase.
iii. The distance between the sources of the wave must be of the order of the wavelength.
Total internal reflection
i. The waves must travel from a denser medium to a less dense medium.
ii. The angle of incidence (in the denser medium) must be greater than the critical angle.
iii. It must involve two media of different optical densities.
Production of beats
1. The amplitude of the two interfering waves should be the same.
ii. The difference between the frequencies of interfering waves should be small.
iii. The two sources of waves must be sounded together.
(b) Calculation of velocity of sound in air
V = f\(\lambda\)
V = 2f(l\(_{2}\) - l\(_{1}\))
= 2 x 256(100 - 35)
= 33280 cm s\(^{-1}\)
= 332.80 m s\(^{-1}\)
(c)(i) Classification of musical instruments
i. Wind
ii. Stringed
iii. Percussion
(ii) Examples of classified instruments saxophone
I. Wind: flute, trumpet, pipe, organ,
iii. Stringed: piano, guitar, violin
iii. Percussion: piano, drum, bell, gong. xylophone
(d) Calculation of critical angle
\(\eta\) = \(\frac{1}{\text {Sin C}}\)
1.5 = \(\frac{1}{\text {Sin C}}\)
Sin C = 0.667
C = 41.80º
(e) Calculation of temperature
\(\frac{V_{1}}{\sqrt T_{1}} = \frac{V_{2}}{\sqrt T_{2}}\)
\(\frac{240} {\sqrt 375} = \frac {310} {\sqrt T_{2}}\)
240 \(\sqrt T_{2}\) = 310 x 19.36
\(\sqrt T_{2}\) = 625 K = 352 ºC