Friday, April 24, 2015

Physics 9702 Doubts | Help Page 121

  • Physics 9702 Doubts | Help Page 121

Question 611: [Attenuation]
(a) State what is meant by attenuation of a signal.

(b) Transmission cable has length of 30 km. Attenuation per unit length of the cable is 2.4 dB km−1.
Calculate, for a signal being transmitted along the cable,
(i) total attenuation, in dB
(ii) ratio of input power of signal to output power of signal

(c) By reference to answers in (b), suggest why the attenuation of transmitted signals is usually expressed in dB

Reference: Past Exam Paper – June 2014 Paper 41 & 43 Q14

Solution 611:
(a) The attenuation of a signal is the reduction in power of the signal

{The attenuation per unit length of the cable is 2.4 dB km−1. So, in 1km of cable, the signal is attenuated by 2.4 dB. For 30km of cable, attenuation = 30 (2.4) dB.}
Total attenuation = 30 (2.4) = 72dB

{The total attenuation is 72 dB. So, the power of the output signal has been reduced by 72 dB when compared to the power of the input signal. [This is the meaning of attenuation as stated before.]
But attenuation is also given by Gain / attenuation / dB = 10 lg(P2/P1).}
Gain / attenuation / dB = 10 lg(P2/P1)
{The general formula is Gain / dB = 10 lg(P2/P1). Note that this is only a formula for comparing 2 quantities, here P2 and P1.

If there is a gain, POUT should be greater than PIN. (This is not the case here)

If there is an attenuation, POUT is less than PIN (this is the case here). So, the ratio POUT / PIN would be a fraction with a value between 0 and 1. The log of such a value is negative. (What I have been explaining now is maths, not physics.) So, we should also take the value of 72 with a negative sign.

Like I said, the formula is only comparing 2 quantities. Let’s say we say want the ratio PIN / POUT. The value of this ratio would be greater than 1 since PIN is greater than POUT. The log would give a positive value. That’s why we take 72 with a positive sign here.}
72 = 10 lg(PIN/POUT)                or – 72 = 10 lg(POUT/PIN)
{Consider 72 = 10 lg(PIN/POUT)
lg(PIN/POUT) = 72 / 10}
PIN/POUT = 107.2 = 1.6x107}
Ratio (= PIN/POUT) = 1.6x107  

(c) Example
It enables smaller / more manageable numbers to be used
The gains in dB for series amplifiers are added, not multiplied

Question 612: [Current of Electricity]
Cell has electromotive force (e.m.f.) E and internal resistance r. It is connected in series with a variable resistor R, as shown in Fig.1.

(a) Define electromotive force (e.m.f.)

(b) Variable resistor R has resistance X. Show that
Power dissipated in resistor R / power produced in cell = X / X+R

(c) Variation with resistance X of the power PR dissipated in R is shown in Fig.2.

(i) Use Fig.2 to state, for maximum power dissipation in resistor R, magnitude of this power and the resistance of R.
(ii) Cell has e.m.f. 1.5 V.
Use answers in (i) to calculate internal resistance of the cell.

(d) In Fig.2, it can be seen that, for larger values of X, the power dissipation decreases. Use relationship in (b) to suggest one advantage, despite the lower power output, of using the cell in a circuit where the resistance X is larger than the internal resistance of the cell.

Reference: Past Exam Paper – November 2009 Paper 21 Q6

Solution 612:
(a) Electromotive force (e.m.f.) is the energy transferred from a source / changed from some form to electrical per unit charge (to drive a charge round a complete circuit).

Power in resistor R = I2X
{Ohm’s law: V = IR. Total resistance = X+r}
And E = I (X+r)
Power in cell = EI and algebra clear leading to ratio = X / (X+r)
{Power in R / power in cell = I2X / EI = IX / E
From E = I (X+r), current I = E / (X+r)
Power in R / power in cell = [E / (X+r)] X / E = X / (X+r)}

Maximum power = 1.4W
Resistance = 0.40Ω                 (allow ± 0.05Ω)

{Power, P = I2R so, I = √(P/R)}
Current in the circuit = (√(P/R) =) √(1.4/0.4) = 1.87A
{E = I (X+r)}
1.5 = 1.87 (0.40 + r)
Internal resistance, r = 0.40Ω

{Power dissipated = I2X = [E / (X+r)]2X = E2X / (X+r)2.
Power in the cell = EI = E2 / (X+r).
From the equation obtained for the power dissipated, for a greater value of X, the power dissipated is less since the denominator is squared. So, less power is lost (there is greater efficiency).}
EITHER Less power is lost / energy wasted / lost OR There is greater efficiency (of energy transfer)

Question 613: [Dynamics > Resultant force]
Graph shows how total resistive force acting on a train varies with its speed.

Part of this force is due to wheel friction, which is constant. The rest is due to wind resistance.
What is the ratio (wind resistance / wheel friction) at a speed of 200 km h–1?
A 4                              B 5                              C 8                              D 10

Reference: Past Exam Paper – November 2013 Paper 13 Q16

Solution 613:
Answer: A.
The wheel friction is constant and the wind resistance increases with speed.

When the speed is zero, the wind resistance is also zero. So, wheel friction = 8 kN.

At a speed of 200 km h–1, the total resistive force is 40 kN.
Wheel friction = 8 kN
Wind resistance = 40 – 8 = 32 kN

Ratio = wind resistance / wheel friction = 32 / 8 = 4

Question 614: [Dynamics]
(a) State Newton’s first law.

(b) A log of mass 450 kg is pulled up a slope by a wire attached to a motor, as shown in Fig.1.

Angle that the slope makes with the horizontal is 12°. Frictional force acting on the log is 650 N. The log travels with constant velocity.
(i) With reference to motion of the log, discuss whether the log is in equilibrium
(ii) Calculate tension in wire
(iii) State and explain whether the gain in potential energy per unit time of the log is equal to output power of the motor

Reference: Past Exam Paper – June 2012 Paper 22 Q3

Solution 614:
(a) Newton’s first law of motion states that a body continues at rest or constant velocity unless acted on by a resultant (external) force.

(i) The log moves with constant velocity (zero acceleration), so no resultant force acts on it. Since there is no resultant force (and no resultant torque), the log is in equilibrium.

{Weight W of the log = mg = 450 (9.81)
From the laws of angles, the weight (which is vertically downwards) is at an angle of 12° to the line perpendicular to the slope.
So, the component of the weight along the slope is W sin12.}
Component of weight of the log acting down the slope = (450)(9.81)sin(12) (=917.8)

{That component of the weight acts down along the slope. The log is being pulled up along the slope by the motor. So, the direction of motion of the log is up along the slope while the component of the weight discussed above is down along the slope. Friction, which always opposes motion, will also be down along the slope since the log is moving up along the slope.
For zero acceleration, the resultant force along the slope should be zero.
Upwards forces along slope = Downwards forces along slope
Tension = Friction + Component of weight}
Tension = 650 + 450gsin(12) = (650 + 917.8)
Tension = 1600 (1570) N

(iii) Work is done against frictional force (friction between log and slope). So, the output power is greater than the gain in potential energy per unit time.

Question 615: [Ultrasound]
(a) By reference to ultrasound waves, state what is meant by specific acoustic impedance of a medium.

(b) A parallel beam of ultrasound of intensity I is incident normally on a muscle of thickness 3.4 cm, as shown in Fig.1.

The ultrasound wave is reflected at a muscle-bone boundary. Intensity of the ultrasound received back at the transducer is IR.
Some data for bone and muscle are given in Fig.2.
specific acoustic impedance / kg m−2 s−1         linear absorption coefficient / m−1
bone                            6.4 × 106                                              130
muscle                         1.7 × 106                                              23

(i) Intensity reflection coefficient α for two media having specific acoustic impedances Z1 and Z2 is given by
α = (Z1 − Z2)2 / (Z1 + Z2)2.
Calculate fraction of the ultrasound intensity that is reflected at the muscle-bone boundary.

(ii) Calculate fraction of the ultrasound intensity that is transmitted through a thickness of 3.4 cm of muscle.

(iii) Use answers in (i) and (ii) to determine the ratio IR / I

Reference: Past Exam Paper – November 2014 Paper 43 Q11

Solution 615:
(a) The specific acoustic impedance of a medium is the product of the density of the medium and the speed of the wave in the medium

(i) Intensity reflection coefficient α = (6.4 – 1.7)2 / (6.4 + 1.7)2 = 0.34

(ii) I / I0 = e–μx = exp (–23 × 3.4 × 10–2) = 0.46

{A fraction of the intensity is lost on transmission and a fraction is lost on reflection. There are two transmission losses and one reflection loss (the ultrasound is transmitted, reflected at the boundary, and finally transmitted back), and each loss is cumulative, so it is necessary to multiply the fractions together.
So, we consider I / I0 twice – that is, (I / I0)2 and α once. We multiply them all.}
Ratio of IR / I (= [(I / I0]2 × α) = (0.46)2 × 0.34 = 0.072

Question 616: [Simple Harmonic motion]
(a) Define simple harmonic motion.

(b) A horizontal plate is vibrating vertically, as shown in Fig.1.

Plate undergoes simple harmonic motion with a frequency of 4.5 Hz and amplitude 3.0 mm.
A metal cube of mass 5.8 g rests on the plate.
Calculate, for the cube, the energy of oscillation.

(c) Amplitude of oscillation of the plate in (b) is gradually increased. The frequency remains constant.
At one particular amplitude, the cube just loses contact momentarily with the plate.
(i) State position of the plate in its oscillation at the point when the cube loses contact.
(ii) Calculate this amplitude of oscillation.

Reference: Past Exam Paper – November 2011 Paper 43 Q3

Solution 616:
(a) In simple harmonic motion, the acceleration is proportional to the displacement / distance from fixed point and is in opposite directions / directed towards fixed point.

{The cube is a temporarily at rest when the displacement is maximum (amplitude)}
Energy of oscillations at rest position = ½ mω2x02     and ω = 2πf
Energy of oscillations = ½ × (5.8 × 10–3) × (2π × 4.5)2 × (3.0 × 10–3)2 = 2.1 × 10–5 J

{The reaction force between the cube and the plate is a minimum when the plate is at its maximum displacement above its mean position. This will, therefore, be the position of the plate when the cube just loses contact.}
The cube loses contact at the maximum displacement above the rest position.

{As stated previously, at the maximum displacement, the plate is temporarily at rest. The plate would be at its maximum displacement when its acceleration is 9.81 m s–2. This is because the acceleration of free fall, g, is always present. The plate has reached its maximum displacement above the mean position, so it won’t be moving any more up. Actually, it is temporarily at rest because the downward acceleration due to gravity has reduced its upwards velocity to zero.}
Acceleration = (–)ω2x0 and acceleration = 9.81 or g
9.81 = (2π × 4.5)2 × x0
Amplitude x0 = 1.2 × 10–2 m


  1. for solution 612, how did you get the power off the cell as EI, did you use the formula P=VI? And if you did, why can't we use the same formula to find the power of the resistor R but instead its I^2R

    1. Power = VI
      For the cell, the p.d. V across its terminal is equal to the e.m.f. E of the cell.

      For a resistor, V is the p.d. across the resistor and I is the current through the resistor. From Ohm's law: V = IR. So, we can replace V = IR in the equation P = VI. This gives P = I^2 R

      This symbol for different significance for the different components we are calculating

    2. The symbol HAS different ...

  2. solution of 616 c 2 ,, even if the plate doesnt go up any further how do I know that is the point when the cube loses contact and not any earlier?


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