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.

Baca Juga

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