• Physics 9702 Doubts | Help Page 243

Question 1124: [Electric field]

Two conducting layers of a liquid crystal display of a calculator are 8 μm apart. A 1.5 V cell is connected across the conducting layers when the calculator is switched on.

What is the electric field strength between the layers?

A 1.2 × 10^–5 V m^–1

B 0.19 V m^–1

C 12 V m^–1

D 1.9 × 10⁵ V m^–1

Reference: Past Exam Paper – June 2013 Paper 12 Q30

Solution 1124:

Answer: D.

Electric field strength, E = V / d = 1.5 / (8×10^-6) = 1.9×10⁵ V m^-1

Question 1125: [Simple harmonic motion]

A spring is hung from a fixed point. A mass of 130 g is hung from the free end of the spring, as shown in Fig. 3.1.

Fig. 3.1

The mass is pulled downwards from its equilibrium position through a small distance d and is released. The mass undergoes simple harmonic motion.

Fig. 3.2 shows the variation with displacement x from the equilibrium position of the kinetic energy of the mass.

Fig. 3.2

(a) Use Fig. 3.2 to

(i) determine the distance d through which the mass was displaced initially, [1]

(ii) show that the frequency of oscillation of the mass is approximately 4.0 Hz. [6]

(b) (i) On Fig. 3.2, draw a line to represent the total energy of the oscillating mass. [1]

(ii) After many oscillations, damping reduces the total energy of the mass to 1.0 mJ.

For the oscillations with reduced energy,

1. state the frequency,

2. using the graph, or otherwise, state the amplitude.            [2]

Reference: Past Exam Paper – November 2007 Paper 4 Q3

Solution 1125:

(a)

(i) d = 0.8cm  

{12 small squares after 0.5cm. 20 small squares à 0.5. So, 12 small squares à (12/20) x 0.5 = 0.3cm. So, d = 0.5+0.3 = 0.8cm}

(ii)

{Frequency can be found from angular frequency, which can be calculated from the kinetic energy}

Maximum kinetic energy = 2.56mJ    

v_max = ωa

Maximum kinetic energy = ½ mω²a²              or ½ mω²(a² – x²)

2.56×10^-3 = ½ (0.130) ω² (0.8×10^-2)²

 ω = 24.8 rad s^-1

f = ω / 2π = 4.0Hz (3.95Hz)  

(b) (i) A line parallel to the x-axis at 2.56mJ is drawn

(ii)

1. Frequency: 4.0Hz

2. Using graph, or otherwise, state amplitude: 0.50cm (allow ± 0.03cm)
{Detailed explanations for this part of the question is available as Solution 261 at Physics 9702 Doubts | Help Page 42 - http://physics-ref.blogspot.com/2014/12/physics-9702-doubts-help-page-42.html}

Question 1126: [Diffraction grating]

Which electromagnetic wave phenomenon is needed to explain the spectrum produced when white light falls on a diffraction grating?

A coherence

B interference

C polarisation

D refraction

Reference: Past Exam Paper – June 2011 Paper 11 Q26 & Paper 13 Q25

Solution 1126:

Answer: B.

When light falls on a diffraction grating, it is diffracted at each slit. The diffracted waves from the slit interference both constructively and destructively to produce a pattern of bright and dark fringes.

Refraction is the bending of light and is different from diffraction.