# Physics 9702 Doubts | Help Page 187

__Question 912:__

__[Oscillations > Simple harmonic motion]__
A student sets out to investigate
the oscillation of a mass suspended from the free end of a spring, as
illustrated in Fig.1.

The mass is pulled downwards and
then released. The variation with time t of the displacement y of the mass is
shown in Fig.2.

**(a)**Use information from Fig.2

(i) to explain why the graph
suggests that the oscillations are undamped,

(ii) to calculate angular frequency
of the oscillations,

(iii) to determine maximum speed of
the oscillating mass.

**(b)**

(i) Determine resonant frequency f

_{0}of the mass-spring system.
(ii) The student finds that if short
impulsive forces of frequency ½ f

_{0}are impressed on the mass-spring system, a large amplitude of oscillation is obtained. Explain this observation.**Reference:**

*Past Exam Paper – November 2002 Paper 4 Q3*

__Solution 912:__**(a)**

(i) It has a constant amplitude.

(ii)

Period T = 0.75s

Angular frequency ω = 2π / T
= 8.4 rad s

^{–1}
(iii)

EITHER use of gradient OR v = ωy

_{0}
{y

_{0}is the amplitude}
Maximum speed v = {8.4 × 0.02 =} 0.168 m s

^{–1}**(b)**

(i) Resonant frequency f

_{0}{= 1/T = 1/0.75} = 1.3Hz
(ii) At ½ f

_{0}, a ‘pulse’ is provided to the mass on alternate/some oscillations. So, the ‘pulses’ build up the amplitude.

__Question 913: [Current of Electricity]__**(a)**Distinguish between

*potential difference*(p.d.) and

*electromotive force*(e.m.f.) in terms of energy transformations.

**(b)**Two cells A and B are connected in series with a resistor R of resistance 5.5 Ω, as shown in Fig.1.

Cell A has e.m.f. 4.4 V and internal
resistance 2.3 Ω. Cell B has e.m.f. 2.1 V and internal resistance 1.8 Ω.

(i) State Kirchhoff’s second law.

(ii) Calculate current in the
circuit.

(iii) On Fig.1, draw an arrow to
show the direction of the current in the circuit. Label this arrow I.

(iv) Calculate

1. p.d. across resistor R,

2. terminal p.d. across cell A,

3. terminal p.d. across cell B.

**Reference:**

*Past Exam Paper – November 2011 Paper 23 Q4*

__Solution 913:__**(a)**

Potential difference (p.d.) is the
energy transformed from electrical to other forms, per unit charge.

Electromotive force (e.m.f.) is the energy
transformed from other forms to electrical per unit charge.

**(b)**

(i) Kirchhoff’s second law states
that the sum of e.m.f.s (in a closed circuit) is equal to the sum of potential
differences.

(ii)

4.4 – 2.1 = I × (1.8 + 5.5 + 2.3)

Current I = 0.24 A

(iii) The arrow (labelled) I should
be shown anticlockwise.

(iv)

1. V = I × R = 0.24 × 5.5 = 1.3(2) V

2. V

_{A}= 4.4 – (I × 2.3) = 3.8(5) V
{Voltage loss in cell A = I
× 2.3}

3.

EITHER V

_{B}= 2.1 + (I × 1.8) OR V_{B}= 3.8 – 1.3
{Here, we need to add (I ×
1.8) to the e.m.f. of cell B because the current is flowing towards the
positive terminal of cell B.}

V

_{B}= 2.5(3) V

__Question 914: [Dynamics > Moments]__
Which pair of forces acts as a
couple on the circular object?

**Reference:**

*Past Exam Paper – November 2008 Paper 1 Q14*

__Solution 914:__**Answer: A.**

A couple consists of 2 forces, equal
in magnitude. But acting in opposite directions, so that they produce a
resultant turning effect. Thus, these 2 forces produces a resultant moment,
while the resultant force is zero.

Choice B: both forces are in the same
direction + their magnitudes are not equal

Choice C: different magnitudes

Choice D: directions are not
opposite to each other

__Question 915: [Kinetic theory of Gases]__**(a)**

(i) State the basic assumption of
the kinetic theory of gases that leads to the conclusion that potential energy
between the atoms of an ideal gas is zero.

(ii) State what is meant by the

*internal energy*of a substance.
(iii) Explain why an increase in
internal energy of an ideal gas is directly related to a rise in temperature of
the gas.

**(b)**A fixed mass of an ideal gas undergoes a cycle PQRP of changes as shown in Fig.1.

(i) State the change in internal
energy of the gas during one complete cycle PQRP.

(ii) Calculate work done on the gas
during the change from P to Q.

(iii) Some energy changes during the
cycle PQRP are shown in Fig.2.

Complete Fig.2 to show all of the
energy changes.

**Reference:**

*Past Exam Paper – November 2010 Paper 41 & 42 Q2*

__Solution 915:__**(a)**

(i) There are no forces (of attraction
or repulsion) between the atoms / molecules / particles.

(ii) The internal energy of a substance
is the sum of kinetic and potential energy of the atoms / molecules due to
their random motion.

(iii) The (random) kinetic energy
increases with temperature. There is no potential energy. (since the gas is
ideal) So, an increase in temperature increases the internal energy.

**(b)**

(i) Change in internal energy = Zero

(ii) Work done = pΔV = (4.0x10

^{5})(6x10^{-4}) = 240J
(iii)

{ΔQ = ΔU – ΔW

+ΔQ is the amount of
heat/energy supplied to the gas.

+ΔU is the increase in
internal energy.

ΔW is the work done ON the
gas. (If + ΔW was used, it would have
been ‘work done BY gas’ – but this is not what we are calculating here. The
table label is ‘work done ON gas’ – so, we used a negative sign in the equation
to account for positive values in the table’s numerical values for this column.)

ΔW = p ΔV

For P to Q: ΔQ = - 600 J. As calculate above, ΔW
= + 240 J. ΔU = - 600 + 240 = - 360 J

For Q to R: ΔQ = +720 J. ΔV = 0, so ΔW = 0. ΔQ = ΔU
– 0 = +720 J

For R to P: ΔQ = +480 J. As answered in (b)(i),
the change in internal energy during one complete cycle is zero. Let the change
in internal energy for this change be ΔU. Consider the changes in internal
energy for P to Q and Q to R. -360 + 720 + ΔU = 0. So, ΔU = -360 J. ΔW = ΔU – ΔQ
= -360 – 480 = -840 J}

For 42/O/N/10 Q.2(b)(iii) How to get the values of the energy changes in the table? Which method should I use?

ReplyDeleteThe method has now been included

DeletePlease consider answering ALL of the following questions before October:

ReplyDelete4/O/N/02 Q.5(b),Q.6(c)(i)

6/O/N/02 Q.11(a)(b)

6/O/N/03 Q.9

04/M/J/04 Q.4(c),Q.8(a),(b)(i),(ii)1.

06/M/J/04 Q.9(b)(iii),Q.11(b)

06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)

04/M/J/05 Q.7(a)

06/O/N/05 Q.8(b),Q.10(a)

04/M/J/06 Q.6(a),(c),Q.7(b)

06/M/J/06 Q.14(b)

04/O/N/06 Q.3(c)

06/O/N/06 Q.3(b)

05/M/J/07 Q.2(d)

04/O/N/07 Q.7(b)(i),(c),Q.10(c)

04/M/J/08 Q.3(c)(ii),Q.5(b),Q.9(b),Q.11(a)(ii)

04/O/N/08 Q.7(c)

04/M/J/09 Q.10(c)(ii),Q.11(b)(iii)

41/O/N/09 Q.6(a),(b)(i),Q.10

42/O/N/09 Q.5(a),Q.7(b)(ii)

41/M/J/10 Q.6(a),Q.7(a)

42/M/J/10 Q.6(a)(ii)

51/M/J/10 Q.2(d)

42/O/N/10 Q.3(c)

For 04/M/J/04 Q.4(c), see solution 916 at

Deletehttp://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-188.html

2004 O/N paper01 Q15

ReplyDeletethe kinetic energy of a particle is increased by a factor of 4,

by what factor does the speed increase?

See solution 67 at

Deletehttp://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-12.html

Hi,

ReplyDeleteWhy is the change in internal energy during one complete cycle in pqrp is zero?

From the first law of thermodynamics, the internal energy depends on the state of the system, that is, its temperature, pressure and volume.

ReplyDeletein a complete cycle, the final state of the system (P, V and T) and the same as its initial state. So, there is no change.

change in internal energy = 0