Question 1
A uniform string is held between a fixed point P and a
variable-frequency oscillator, as shown in Fig. 5.1.
Fig. 5.1
The distance between point P and the oscillator is L.
The frequency of the oscillator is adjusted so that the
stationary wave shown in Fig. 5.1 is formed.
Points X and Y are two points on the string.
Point X is a distance 1/8 L from the end of the string
attached to the oscillator. It vibrates with frequency f and amplitude A.
Point Y is a distance 1/8 L from the end P of the string.
(a) For the vibrations of point Y, state
(i) the frequency (in terms of f ), [1]
(ii) the amplitude (in terms of A).
[1]
(b) State the phase difference between the vibrations of
point X and point Y. [1]
(c) (i) State, in terms of f and L,
the speed of the wave on the string. [1]
(ii) The wave on the string is a stationary
wave.
Explain, by reference to the formation of a stationary
wave, what is meant by the
speed stated in (i). [3]
Reference: Past Exam Paper – November 2009 Paper 22 Q5
Solution 1:
(a)
(i) f [B1]
(ii) A [B1]
{Amplitude
means maximum displacement.}
(b) 180o or π rad (unit necessary) [B1]
{When talking about the phase of a point, we need to
consider its displacement and the direction of motion/vibration.
Phase difference between 2 points means the difference in
phase about the 2 points.
Point X and Y are at the same displacement (one is +ve
and the other -ve). However, they moving in opposite direction. So, they are
out of phase (phase difference = 180°).}
(c) (i) Speed
of wave = f x L [B1]
(ii)
The wave is reflected at the end /
at P. [B1]
EITHER The incident and reflected
waves interfere OR the two waves travelling in opposite directions interfere. [M1]
The speed is the speed of incident
or reflected wave / one of these waves. [A1]
