Question 31
(a)
On Fig. 4.1, complete the two graphs to illustrate what is meant by
the amplitude A, the wavelength λ and the period T
of a progressive wave.
Ensure that you label
the axes of each graph.
Fig. 4.1
[3]
(b)
A horizontal string is stretched between two fixed points X and Y. A
vibrator is used to oscillate the string and produce a stationary wave. Fig.
4.2 shows the string at one instant in time.
Fig. 4.2
The speed of a
progressive wave along the string is 30 m s-1. The stationary wave
has a
period of 40 ms.
(i)
Explain how the stationary wave is formed on the string. [2]
(ii)
A particle on the string oscillates with an amplitude of 13 mm. At
time t, the particle has
zero displacement.
Calculate
1.
the displacement of the particle at time (t
+ 100 ms),
2.
the total distance moved by the particle from time t
to time (t + 100 ms).
[3]
(iii)
Determine
1.
the frequency of the wave, [1]
2.
the horizontal distance from X to Y. [3]
[Total: 12]
Reference: Past Exam Paper – November 2018 Paper 23 Q4
Solution:
(a)
graph with x-axis labelled ‘distance’
and wavelength/λ correctly shown
graph with x-axis labelled ‘time’ and
period/T correctly shown
graph with y-axis labelled
‘displacement’ and amplitude/A correctly shown
(b)
(i)
The wave moves along the string and gets reflected at the
fixed point. The incident and reflected wave interfere to form the stationary
wave.
(ii)
1.
{A particle on the string can only move up and
down.
Let the particle be at a displacement of +13 mm
initially.
In a time equal to the period (= 40 ms), the
particle would have moved and reached its same position again. That is, in 1
period, the particle moves
1. from +13 mm (max position above equilibrium)
to the equilibrium position
2. from equilibrium position to – 13 mm (max
position below equilibrium) [this is
half a period]
3. from + 13 mm back to equilibrium position
4. from equilibrium position back to + 13 mm
Since the period is 40 ms, each of these step
would take 10 ms as after 1 period, the particle should regain its original position.
A time of t + 100 ms means that we want to find
the position after 100 ms.
But first, we need to find out how many periods
are there in 100 ms.}
{Number of periods =} 100 / 40 or 2.5 (cycles/periods/T)
{So in 100 ms, the particle would have
travelled 2 and a half cycle {it has completed two and a half oscillations}.
But we know that after each cycle, the particle regains its original position.
After the 2 complete cycle, the particle is at its original position again.
Half a cycle remains.
From above, in half a cycle (step 2 of 4), the particle
would be at the equilibrium position.}
displacement = 0
2.
{Again, from above, in a time of one period,
the particle has completed these 4 steps. In each step, the particle moves a
distance of 13 mm.
In one period, distance travelled = 13 + 13 + 13 + 13 = 52 mm
In two period, distance travelled = 52 × 2 = 104 mm
In half a cycle, distance travelled = 13 + 13 =
26 mm
Total distance travelled = 104 + 26 = 130 mm}
distance = 130 mm
(iii)
1.
{Frequency
= 1 / period}
f = 1 / (40×10-3)
f = 25 Hz
2.
v = fλ or λ = vT
λ =
30 / 25 or 30 × 40×10-3 (= 1.2 m)
{distance from X to Y is one and a half
wavelength (1.5λ)}
distance = 1.2 × 1.5
distance = 1.8 m
