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Question 621: [Waves > Stationary waves]

(a) State two features of stationary wave that distinguish it from a progressive wave

(b) Long tube is open at one end. It is closed at other end by means of a piston that can be moved along the tube, as shown Fig.1. 

Loudspeaker producing sound of frequency 550Hz is held near the open end of tube.

Piston is moved along the tube and a loud sound is heard when the distance L between piston and open end of tube is 45cm. Speed of sound in tube = 330ms^-1.

(i) Show that wavelength of the sound in the tube is 60cm

(ii) On Fig.1, mark all positions along the tube of:

1. Displacement nodes (label with letter N)

2. Displacement antinodes (label letter A)  

(c) Frequency of the sound produced by loudspeaker in (b) is gradually reduced.

Determine lowest frequency at which a loud sound will be produced in the tube of length L = 45cm

Reference: Past Exam Paper – June 2010 Paper 22 Q4

Solution 621:

(a) Choose any 2:

No energy transfer

The amplitude varies along its length / nodes and antinodes

The neighboring points (in an inter-nodal loop) vibrate in phase, etc.

(b)

(i)

(v = fλ)

Wavelength λ = v / f = 330 / 550 = 0.6m = 60cm

(ii) 1. & 2.

A node labeled at the piston and antinode labeled at the open end of the tube. Additional node and antinode in correct positions along tube.

{The length L represent ¾ of a wavelength (45 / 60 = ¾ ) because there must be an antinode at the open end and a node at the piston. I think that at 100%, L = 9cm (measured using a ruler). The other antinode should be found at ¼ wavelength from the piston. This is 9 / 3 = 3cm from the piston. The other node is at ½ wavelength, that is a distance 2x3 = 6cm from the piston.}

(c)

At lowest frequency, length of tube corresponds to λ /4

So, λ = 4x45 = 180cm = 1.8m

Frequency = (v/ λ =) 330 / 1.8 = 180Hz

Question 622: [Current of Electricity > Resistance]

Resistance of a metal cube is measured by placing it between two parallel plates, as shown.

Cube has volume V and is made of a material with resistivity ρ. Connections to the cube have negligible resistance.

Which expression gives electrical resistance of the metal cube between X and Y?

A ρV^1/3                        B ρV^2/3                                    C ρ / V^1/3                     D ρ / V^2/3

Reference: Past Exam Paper – June 2011 Paper 13 Q35

Solution 622:

Answer: C.

Resistance R of a wire = ρL / A

Where L is the length of the wire and A is the cross-sectional area of the wire.

For a cube, all the sides are of equal lengths. Let the length be L/

Volume V = L³

Length = L and cross-sectional area (in contact with the plates) = L²

Resistance R = ρL / L² = ρ / L

But since V = L³, length L = V^1/3

Resistance R = ρ / V^1/3

Alternatively, if we consider the units of the quantities involved, only choice C gives the unit of resistance (Ω).

Unit of resistivity = Ωm

Unit of volume V = m³

Unit of ρ / V^1/3 = [Ωm] / [m³]^1/3 = [Ωm] / m = Ω   

Question 623: [Quantum Physics]

(a) State an effect, one in each case, that provides evidence for

(i) wave nature of a particle

(ii) particulate nature of electromagnetic radiation

(b) Four electron energy levels in an atom shown in Fig.1. 

An emission spectrum is associated with electron transitions between these energy levels.

For spectrum

(i) state number of lines

(ii) calculate minimum wavelength

Reference: Past Exam Paper – November 2010 Paper 41 & 42 Q7

Solution 623:

(a)

(i) e.g. electron / particle diffraction  

(ii) Photoelectric effect

(b)

(i) Number of lines: 6

{The question did not ask for visible lines only. So, we include all the lines even if they are outside the visible range.}

(ii)

(Emission spectrum is due to transition from higher energy level to lower one)

(Energy E = hf = hc / λ. So the wavelength is minimum when the change in energy is biggest.)

Change in energy (= [-0.87x10^-19] – [-5.44x10^-19]) = 4.57x10^-19J

Wavelength λ = hc / E = [(6.63x10^-34)(3.0x10⁸)] / (4.57x10^-19) = 4.4x10^-7m 

Question 624: [Graphs]

(a) On axes of Fig.1, sketch variation with distance from a point mass of the gravitational field strength due to the mass.

(b) On axes of Fig.2, sketch variation with speed of the magnitude of the force on a charged particle moving at right-angles to a uniform magnetic field.

(c) On axes of Fig.3, sketch variation with time of the power dissipated in a resistor by a sinusoidal alternating current during two cycles of the current.

Reference: Past Exam Paper – November 2014 Paper 43 Q2

Solution 624:

(a)

{Gravitational field strength, g = GM / x²}

The graph is a smooth curve with decreasing gradient, not starting at x = 0.

{At zero distance from the point mass, the gravitational field strength is infinite.}

The end of the line should not be not at g = 0 or horizontal. {The gradient is not zero.}

(b)

{F = Bqv. The force F depends linearly on the speed v. Charge q and magnetic field B are constant.}

The graph is a straight line with positive gradient. The line starts at the origin.

(c)

{Power P dissipated = I²R. An alternating current is used, so the current can be described as varying sinusoidally. I = I₀ sin(ωt).

However, in the formula, we take the square of the current. So, all the negative values would become positive – that is, the graph is a sin² curve.}

The graph has a sinusoidal shape with only positive values and peak / trough are at height constant. At least, 4 ‘loops’ should be shown.

Question 625: [Waves > Interference > Double Slit]

Diagram shows a view from above of double slit interference demonstration.

L is monochromatic light source with a vertical filament. B is a barrier with two narrow vertical slits and S is screen upon which interference fringes form.

Intensity is I at a point on the screen where centre of the fringe pattern forms.

What is the intensity, at the same point, when one of the slits is covered up?

A I / √2                       B I / 2                          C I / 2√2                                  D I / 4

Reference: Past Exam Paper – November 2011 Paper 11 Q28 & Paper 13 Q27

Solution 625:

Answer: D.

At the central fringe, constructive interference occurs, i.e. the resultant displacement is the sum of displacements of each wave (here there is 2 waves due to the 2 slits).

By covering up one slit, the wave amplitude is halved because only 1 wave is reaching that point.

Since intensity is proportional to (amplitude)², the intensity is one-quarter of the original intensity when the amplitude is halved..