Q#22 (Past Exam Paper – June 2018 Paper 12 Q28)
The diagram shows a stationary wave, at time t = 0, that has been set up on a string fixed between points P and S.
The nodes of the stationary wave occur on the string at P, Q, R and S. Point X is moving down at time t = 0. The points on the string vibrate with time period T and maximum amplitude 2 cm.
The displacement s is positive in the upward direction.
Which graph best shows the variation with t of the displacement s of point Y on the string?
Solution:
Answer: A.
Loops would be formed between the nodes PQ, QR and RS.
Vibrations in adjacent loops are out of phase. That is, if the vibration is downwards at X (in the PQ loop), then vibrations in the loop QR would be upwards (out of phase with the loop PQ).
So, vibrations in loop RS would also be out of phase with loop QR as they are adjacent. Since vibrations in loop QR is upwards, it means that vibration is loop RS will be downwards.
So, the vibrations at points X and Y are in phase. The pattern of movement at Y will therefore be the same as at X. The displacement at Y will also be downwards – it has a negative displacement. [C and D are incorrect]
However, the maximum amplitude at Y will be lower than at X because X is at an antinode and Y is not. [A is correct]
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