(a)(i) What is a wave motion?
(ii) State two differences between a radio wave and a sound wave.
(b)(i) Given that you are provided with a tuning fork, a burette and other necessary apparatus, describe with the aid of a diagram, an experiment to determine the frequency of a note emitted by a source of sound. [assume the velocity of sound in air is known]
(ii)State two precautions necessary to obtain accurate result in the experiment described in (b)(i) above
(c) A pipe closed at one end is 100 cm long. If the air in the pipe is set into vibration and a fundamental note is produced, calculate the frequency of the note. [ velocity of sound in air = 340 ms\(^{-1}\)]
Explanation
(a)(i) A wave motion is a process of transferring a disturbance in form of K.E from one point to another in a medium without any transfer of the particles of the medium.
(ii)
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Radio wave |
Sound wave |
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It is electro-magnetic wave. It is a transverse wave Requires no medium for propagation. Has velocity equal to that of light. |
It is a mechanical wave It is longitudinal wave
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(b) (i)
Fill the burette with water. Set the prongs of the fork into vibration and bring it near the open end of the burette, then gradually open the tap to run out some water, there-by increasing the length of the air column until a loud sound is heard. Measure and record the length L\(_2\) of the air column. Further run out the water while le vibrating tuning fork is still held over open end of the burette until a second loud sound is heard, then read and record the length L\(_2\) of the air column. Last repeat the experiment to obtain L L\(_1\) and L L\(_2\) again and find the mean values of L L\(_1\) and L L\(_2\). Obtain the frequency of the tuning fork using
f = \(\frac{v}{2(L_2 - L_1)}\)
ii) Precautions: - repeat the experiment to obtain nean values
- tap tuning fork on a pad not on a hard surface
- hold tuning fork over the burette to avoid contact
- obtain two lengths of air column to remove rnd correction.
(c) \(f_o = \frac{v}{4L} = \frac{340}{4 \times 100 \times 10^{-2}}\) = 851Hz