Question 20
(a)
State the conditions required for the formation of a stationary
wave. [2]
(b)
A horizontal string is stretched between two fixed points X and Y.
The string is made to vibrate vertically so that a stationary wave is formed.
At one instant, each particle of the string is at its maximum displacement, as
shown in Fig. 4.1.
Fig. 4.1
P and Q are two
particles of the string. The string vibrates with a frequency of 40 Hz.
Distance XY is 2.0 m.
(i)
State the number of antinodes in the stationary wave. [1]
(ii)
Determine the minimum time taken for the particle P to travel from
its lowest point to its highest point. [2]
(iii)
State the phase difference, with its unit, between the vibrations of
particle P and of
particle Q. [1]
(iv)
Determine the speed of a progressive wave along the string. [2]
[Total: 8]
Reference: Past Exam Paper – November 2017 Paper 22 Q4
Solution:
(a)
Stationary waves are formed when (two) waves
travelling (at same speed) but in opposite directions overlap.
The waves (should be of the same type and) have
the same frequency/wavelength.
(b)
(i)
{Antinodes are points of max displacement (e.g.
P)}
5
(ii)
{This
is half a period.}
Period T = 1 / 40 (= 2.5 × 10-2)
s
time taken = 2.5 × 10-2 / 2
= 1.3 × 10-2 s (1.25 × 10-2 s)
(iii)
{For a stationary wave, points within a loop
have the same phase while points in adjacent loops are out of phase by 180°.}
180°
(iv)
v = fλ
{From
the diagram, 2.5 waves corresponds to 20 cm.}
λ = 2.0 / 2.5 (= 0.80 m)
v = 0.80 × 40 =
32 m s-1
Can you please explain (iii)? I don't get what it means by the points within loop. Thanks!
ReplyDeletea loop is formed between 2 nodes. Within these 2 nodes, the particles on the wave would have different displacements. All of these particles are in phase.
DeleteWhen considering a loop just adjacent to another loop, the particles in the 2 loops are out of phase
Is the type of wave transverse or longitudinal
ReplyDeletethe string is horizontal while the vibrations is vertical. So, it is a transverse wave
Delete