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YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Tuesday, January 7, 2020

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.


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

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