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Question 570: [Simple harmonic motion]

A cylinder and piston, used in car engine, are illustrated in Fig.

Vertical motion of the piston in the cylinder is assumed to be simple harmonic.

The top surface of piston is at AB when it is at its lowest position; it is at CD when at its highest position, as marked in Fig.

(a) Displacement d of the piston may be represented by the equation

d = – 4.0 cos(220t)

where d is measured in centimetres.

(i) State the distance between lowest position AB and the highest position CD of the top surface of the piston.

(ii) Determine number of oscillations made per second by the piston.

(iii) On Fig, draw a line to represent the top surface of the piston in the position where the speed of the piston is maximum.

(iv) Calculate maximum speed of the piston.

(b) Engine of a car has several cylinders. Three of these cylinders are shown in Fig.

X is the same cylinder and piston as in Fig.1.

Y and Z are two further cylinders, with lowest and the highest positions of the top surface of each piston indicated.

Pistons in the cylinders each have the same frequency of oscillation, but they are not in phase.

At a particular instant in time, position of the top of the piston in cylinder X is as shown.

(i) In cylinder Y, oscillations of the piston lead those of the piston in cylinder X by a phase angle of 120° (2/3 π rad).

Complete diagram of cylinder Y, for this instant, by drawing

1. line to show the top surface of the piston,

2. arrow to show the direction of movement of the piston.

(ii) In cylinder Z, oscillations of the piston lead those of the piston in cylinder X by a phase angle of 240° (4/3 π rad).

Complete diagram of cylinder Z, for this instant, by drawing

1. line to show the top surface of the piston,

2. arrow to show the direction of movement of the piston.

(iii) For piston in cylinder Y, calculate its speed for this instant.

Reference: Past Exam Paper – November 2010 Paper 43 Q3

Solution 570:

(a)

(i)

{In the equation, 4.0 represents the amplitude – that is the maximum or miminum distance from the equilibrium position. The distance between AB (minimum) and CD (maximum) is twice the amplitude.}

Distance = 8.0cm

(ii)

{The argument of the cosine in the equation is ωt. ω = 2πf = 220.

Number of oscillations per second = frequency}

2πf = 220

Frequency f = 35

(iii) {In simple harmonic motion, the speed is maximum at the equilibrium position.}

The line should be drawn midway between AB and CD

(iv) Maximum speed v = ωa = 220 × 4.0 = 880 cm s^–1

(b)

(i)

{360° corresponds to a period / wavelength – so in 360°, the piston returns to its original position. The piston is originally at AB, its lowest position. The motion of the piston for 1 period is as follows: it moves from AB upwards towards CD and then moves from CD downwards towards AB until it reaches its original position at AB. This whole motion corresponds to 360°.

Thus, CD is halfway in the motion and so will correspond to 180°. From its original position (AB) to CD (180°), the piston moves upwards. Also, when returning from CD to AB (360°), the motion of the piston is downwards.

Now, to find the position of the top of the piston, we need to consider the equation for its displacement

d = – 4.0 cos(220t)

To account for a phase ϕ, the equation is changed to

d = – 4.0 cos(220t + ϕ)

Note that the equation gives the displacement of the piston form the equilibrium position, which is halfway between AB and CD. It varies from -4 to +4 with a negative value being lower than the equilibrium position and a positive value being above the equilibrium position.

But here is an important thing to note. In the previous parts, we identified the distance between AB and CD to be 8.0cm. However, if we measure the distance with a ruler at 100% the size, we will notice that the distance measured is 4cm. That is, the diagram is not drawn to full scale. 1cm (measured by a ruler) represents a displacement of 2cm for the piston. [1cm: 2cm displacement]

Now, when the phase difference is 2π/3 rad, the displacement is [put t = 0]

d = – 4.0 cos(220t + ϕ) = – 4.0 cos(0 + 2π/3) = 2cm

Since the value is positive, the piston is 2cm above the equilibrium and hence, (4+2 =) 6cm above AB. But since the scale is [1cm: 2cm displacement], on the actual diagram at 100%, the line should be drawn 3cm above AB as the diagram is not drawn to full scale.

Similarly, for a phase difference of 4π/3 rad, the displacement is also obtained to be +2.0cm but now, the piston is moving downwards as explained above.}

1. The line should be drawn 3 cm above AB (allow ±2 mm)

2. The arrow should be pointing upwards

(ii)

1. The line drawn 3 cm above AB (allow ±2 mm)

2. The arrow should be pointing downwards

(iii)

{Displacement x = 2.0cm}

Speed v = ω√(a² – x²) = 220 × √(4.0² – 2.0²) = 760 cm s^–1

Question 571: [Current of Electricity]

(a) Define potential difference (p.d.).

(b) Battery of electromotive force 20 V and zero internal resistance is connected in series with two resistors R₁ and R₂, as shown in Fig.1.

Resistance of R₂ is 600 Ω. Resistance of R₁ is varied from 0 to 400 Ω.

Calculate

(i) maximum p.d. across R₂,

(ii) minimum p.d. across R₂.

(c) Light-dependent resistor (LDR) is connected in parallel with R₂, as shown in Fig.2.

When light intensity is varied, the resistance of the LDR changes from 5.0 kΩ to 1.2 kΩ.

(i) For maximum light intensity, calculate the total resistance of R₂ and the LDR.

(ii) Resistance of R₁ is varied from 0 to 400 Ω in the circuits of Fig.1 and Fig.2. State and explain the difference, if any, between minimum p.d. across R₂ in each circuit. Numerical values are not required.

Reference: Past Exam Paper – June 2013 Paper 23 Q6

Solution 571:

(a) Potential difference (p.d.) is defined as the work done / energy transformed (from electrical to other forms) per unit charge.

(b)

(i)

{The p.d. is maximum when R₁ = 0.}

Maximum p.d. = 20 V

(ii)

{p.d. is minimum when R₁ is maximum.}

Minimum p.d. = [600 / (600+400)] x 20 = 12 V

(c)

(i)

{When the light intensity is maximum, the resistance of the LDR is minimum since resistance decreases with an increase in light intensity.}

We need to use the resistance of 1.2 kΩ for the LDR.

1 / R = 1/1200 + 1/600

Total resistance R = 400 Ω

(ii) The total (combined) resistance of the parallel combination of the LDR and R₂ is less than the resistance of R₂ (as in the 1^st case). So, the minimum p.d. is reduced.

{Let the combined resistance of LDR and R₂ be R_c.

p.d. across R_c = [R_c / (R_c+R₁)] x 20

Since R_c is less than R₂ (as in the 1^st case), the p.d. across it is less.}

Question 572: [Dynamics > Momentum]

An object of mass 20 kg is travelling at constant speed of 6.0 m s^–1.

It collides with an object of mass 12 kg travelling at constant speed of 15 m s^–1 in the opposite direction. The objects stick together.

What is the speed of the objects immediately after collision?

A 1.9 m s^–1                  B 9.0 m s^–1                  C 9.4 m s^–1                  D 21 m s^–1

Reference: Past Exam Paper – November 2011 Paper 12 Q11

Solution 572:

Answer: A.

From the law of conservation of momentum, the sum of momentum before the collision should be equal to the sum of momentum after collision.

Before the collision, the 2 objects are moving in opposite directions. So, one of the velocities should be taken as negative since momentum (= mv) is a vector quantity.

Sum of momentum before collision: 20(6) + 12(-15) = -60 kg m s^–1

After the collision, the 2 objects stick together. Let their combined mass be M and their velocity be V.

M = 12+20 = 32 kg

Considering only the magnitudes,

MV = 60

Speed V = 60 / 32 = 1.875 = 1.9ms^-1