# Physics 9702 Doubts | Help Page 112

__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√(a

^{2}– x^{2}) = 220 × √(4.0^{2}– 2.0^{2}) = 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 R

_{1}and R

_{2}, as shown in Fig.1.

Resistance of R

_{2}is 600 Ω. Resistance of R_{1}is varied from 0 to 400 Ω.
Calculate

(i) maximum p.d. across R

_{2},
(ii) minimum p.d. across R

_{2}.**(c)**Light-dependent resistor (LDR) is connected in parallel with R

_{2}, 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 R

_{2}and the LDR.
(ii) Resistance of R

_{1}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 R_{2}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
R

_{1}= 0.}
Maximum p.d. = 20 V

(ii)

{p.d. is minimum when R

_{1}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 R

_{2}is less than the resistance of R_{2}(as in the 1^{st}case). So, the minimum p.d. is reduced.
{Let the combined resistance
of LDR and R

_{2}be R_{c}.
p.d. across R

_{c}= [R_{c}/ (R_{c}+R_{1})] x 20
Since R

_{c}is less than R_{2}(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}
for solution 571 i really don't understand why the p.d must decrease if the total parallel resistance is reduced. because if resistance decrease the current flowing would increase and the p.d must increase in this case?? please reply ASAP.

ReplyDeleteWe know that connecting in parallel reduces the effective resistance. That is Rc, as defined above, is less than R2.

DeleteEffective resistance means that the part of ‘parallel combination of the LDR and R2’ can be replaced by a single resistor Rc. And Rc is in series with R1.

For such a series combination, the same current flows and we can apply the potential divider equation, which is independent of the current. So, we do not need to consider the same in the same.

You have a point, decreasing resistance increases current. We can work out the answer from that viewpoint. we already found the effective resistance of the LDR and R2 which is 400. Using E=sum of IR, we can then calculate the current flowing through the curcuit. I=20/(400+400)=0.25A. As you deduced it does increase using this current we can now calculate the p.d across the parallel network consisting of LDR+R2 , since their combined resistance is 400. Using V=IR, the V=0.25*400=10V , since in LDR and R2 are in parallel construct. So this means the p.d across LDR=p.d across R2=10V. Keeping in mind this is the minimum p.d across

ReplyDeleteR2 we calculated we can see that it is smaller than that of the original circuit in fig.1. However Question pointed out their is no need for numerical answers. But I hope this clears out your problem.