Question 3
A stationary wave is
set up on a stretched string.
The diagram shows the
string at two instants of time when it has maximum displacement.
The oscillations of
point P on the string have amplitude A.
What is the distance
moved by P from the position shown in the diagram after half a time period of
the wave?
A 0 B A C 2A D 4A
Reference: Past Exam Paper – June 2016 Paper 12 Q 25
Solution:
Answer:
C.
Consider the particle at point
P.
After a time equal to the
period T, the particle at point P will again be a point P again.
To complete the period T,
we can equal the time taken into 4 (i.e. we can consider the position of the
particle at time = 0, ¼ T, ½ T, ¾ T and T).
The amplitude A is
measured from the equilibrium position to the maximum displacement (e.g. at
point P).
At
time t = 0, the particle is at point
P.
At
time t = ¼ T, the
particle is at the equilibrium position.
It has moved a distance A.
The spring is straight.
At
time t = ½ T, the
wave formed is the same as the dashed line.
The particle has again
moved a distance A from the equilibrium position.
So, from time t = 0 to
time t = ½ T, total distance moved by the particle = A + A = 2A.
[C is
correct]
At
time t = ¾ T, the
particle is at the equilibrium position again.
Total distance moved by particle
= 2A + A = 3A
At
time t = T, the particle is at point
P again.
Total distance moved by particle = 3A + A = 4A
Hi there,
ReplyDeleteInitially I was under the impression that because stationary waves do not transfer energy, that the particle P would not be able to be transferred. Two waves travelling in opposite directions would mean that their displacements cancel each other out and that there is no net movement.
the wave does not move right or left, but the particles DO move up and down (vibrate)
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