# Physics 9702 Doubts | Help Page 188

__Question 916: [Oscillations > Simple harmonic motion]__
A vertical spring supports a mass,
as shown in Fig.1.

The mass is displaced vertically
then released. The variation with time t of the displacement y from its mean
position is shown in Fig.2.

A student claims that the motion of
the mass may be represented by the equation

y = y

_{0 }sinωt.**(a)**Give two reasons why the use of this equation is inappropriate.

**(b)**Determine angular frequency ω of the oscillations.

**(c)**The mass is a lump of plasticine. The plasticine is now flattened so that its surface area is increased. The mass of the lump remains constant and the large surface area is horizontal.

The plasticine is displaced
downwards by 1.5 cm and then released.

On Fig.2, sketch a graph to show the
subsequent oscillations of the plasticine.

**Reference:**

*Past Exam Paper – June 2004 Paper 4 Q4*

__Solution 916:__**(a)**

Example:

The amplitude
is not constant OR the wave is damped

It should
be (-)cos, (not sin)

**(b)**

Period T = 0.60 s

Angular frequency, ω = 2π / T = 10.5
rad s

^{–1}(allow 10.4 → 10.6)**(c)**

The sketch should have:

same period

displacement always less

amplitude reducing appropriately

__Question 917: [Matter > Deformation of Solids]__
A spring of unextended length 0.50 m
is stretched by a force of 2.0 N to a new length of 0.90 m.

Variation of its length with tension
is as shown.

How much strain energy is stored in
the spring?

A 0.40 J B 0.80 J C
0.90 J D 1.8 J

**Reference:**

*Past Exam Paper – June 2007 Paper 1 Q19*

__Solution 917:__**Answer: A.**

The graph can be interpreted as
follows: a tension of 2.0 N causes the spring to extend, from its original
length, by (0.9 – 0.5) m.

Strain energy = Area under tension-length
graph

Strain energy = ½ × (2.0 – 0) × (0.9 – 0.5) = 0.40 J

__Question 918: [Kinematics > Terminal velocity]__
Which displacement-time graph best
represents the motion of a falling sphere, the initial acceleration of which
eventually reduces until it begins to travel at constant terminal velocity?

**Reference:**

*Past Exam Paper – June 2009 Paper 1 Q5*

__Solution 918:__**Answer: D.**

The gradient of a displacement-time
graph gives the velocity of the falling sphere.

For a constant terminal velocity,
the gradient of the graph should become constant at the later stage. For the
gradient to be constant, the graph should be a straight line in that region.
Thus, the final section of the graph should be a straight line. [A and B are incorrect]

Terminal velocity is the maximum
velocity reached by the falling sphere. It cannot be zero. A horizontal straight
line indicates that the gradient is zero. [C is
incorrect]

__Question 919: [Alternating Current + Electromagnetism]__**(a)**The mean value of an alternating current is zero.

Explain

(i) why an alternating current gives
rise to a heating effect in a resistor,

(ii) by reference to heating effect,
what is meant by root-mean-square (r.m.s.) value of an alternating current.

**(b)**A simple iron-cored transformer is illustrated in Fig.1.

(i) State Faraday’s law of
electromagnetic induction.

(ii) Use Faraday’s law to explain
why the current in the primary coil is not in phase with the e.m.f. induced in
the secondary coil.

**Reference:**

*Past Exam Paper – November 2014 Paper 41 & 42 Q7*

__Solution 919:__**(a)**

(i)

{Power loss in a resistor =
I

^{2}R.}
EITHER The heating effect in a
resistor ∝ (current)

^{2}. But the square of value of an alternating current is always positive. So there is a heating effect.
OR The current moves in opposite
directions in the resistor during half-cycles. The heating effect is
independent of direction.

(ii) It is that value of the direct
current, producing the same heating effect (as the alternating current) in a
resistor.

**(b)**

(i) Faraday’s law of electromagnetic
induction states that the induced e.m.f. is proportional to the rate of change
of (magnetic) flux (linkage).

(ii) The flux in the core is in
phase with the current in the primary coil. There is an (induced) e.m.f. in the
secondary coil because the coil cuts the flux. The flux and the rate of change
of flux are not in phase.

Please 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.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 42/M/J/10 Q.6(a)(ii), go to

Deletehttp://physics-ref.blogspot.com/2014/10/9702-june-2010-paper-42-43-worked.html

m16 qp12 question No.25

ReplyDeleteSee solution 1120 at

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