# Physics 9702 Doubts | Help Page 106

__Question 540: [Waves]__
Long rope is held under tension
between two points A and B. Point A is made to vibrate vertically and a wave is
sent down the rope towards B as shown in Fig.

Time for one oscillation of point A
on rope is 0.20 s. Point A moves a distance of 80 mm during one oscillation.
Wave on the rope has wavelength of 1.5 m.

**(a)**

(i) Explain the term displacement
for wave on the rope.

(ii) Calculate, for wave on the
rope,

1. amplitude

2. speed

**(b)**On Fig, draw wave pattern on the rope at a time 0.050 s later than that shown.

**(c)**State and explain whether waves on the rope are

(i) progressive or stationary

(ii) longitudinal or transverse

**Reference:**

*Past Exam Paper – November 2013 Paper 21 & 22 Q5*

__Solution 540:__**(a)**

(i) The displacement is the distance
of the (particles of the) rope is (above or below) from the equilibrium / mean
/ rest / undisturbed position

(ii)

1.

{During one oscillation,
point A moves a distance of 80mm as given in the question.

During one oscillation,
point A would return to its original position. Any point would move a distance
= 4 time the amplitude during one oscillation,

**a**.
Remember, as defined above,
the displacement is up and down while the wave travels to the right (in this
case). Amplitude is the maximum displacement form the equilibrium position.

Consider the point A for
example, which is originally on the minima. During one oscillation, it will
move

1. a distance

**a**from its current position at A to the equilibrium position,
2. another distance

**a**from the equilibrium position to reach the maxima (crest),
3. another distance

**a**to move from the maxima (crest) back to reach the equilibrium position again,
4. and finally, another
distance

**a**from the equilibrium position to return to its original position at A. Thus,}
Amplitude = 80/4 = 20mm

2.

Time for one oscillation of point A
on the rope = period, T = 0.20s

Wavelength, λ =1.5m

Velocity, v = (frequency,

**f**)**x (wavelength, λ) =****f**λ
Frequency = 1/period = 1/T = 1/0.2 =
5Hz

Speed v =

**f**λ = 5 x 1.5 = 7.5ms^{-1}**(b)**0.05s corresponds to ¼ of the period. Thus the wave has moved ¼ forward (to the right)

Point A should now be shown at the
equilibrium position.

The drawing should have the same
wavelength, shape and the peaks /waves have moved to the right by ¼ λ. The
drawing resembles a (–sine) curve with the point A having moved vertically up
to the equilibrium position, with no movement of A in the horizontal direction.

**(c)**

(i) The wave is progressive since
energy / peaks / troughs is/are transferred / moved / propagated (by the
waves). It cannot be stationary since wave has not reach end B.

(ii) The wave is transverse as the
particles/rope movement is perpendicular to the direction of travel /
propagation of the energy/wave velocity.

__Question 541: [Units]__
An ion is accelerated by a series of
electrodes in vacuum. Graph of power supplied to the ion is plotted against
time.

What is represented by area under
the graph between two times?

A the change in kinetic energy of
the ion

B the average force on the ion

C the change in momentum of the ion

D the change in velocity of the ion

**Reference:**

*Past Exam Paper – June 2010 Paper 11 Q3 & Paper 12 Q4 & Paper 13 Q7*

__Solution 541:__**Answer: A.**

A graph of power against time is
plotted. The area under the graph should correspond to a quantity having the
same units as the product of units of power and time.

Unit of power: W = J/s

Unit of product of power and time: (J/s)
(s) = J

Thus, the area under the graph
represents a quantity having unit ‘Joule’. This is the unit of energy. So, the
area under the graph here represents the change in kinetic energy of the ion.

__Question 542: [Measurement > CRO]__
Oscilloscope display consists of two
separate traces, a waveform and a long horizontal line. Horizontal line may be
taken as the zero level.

Grid on the screen is calibrated in
cm squares, timebase setting is 2.5 ms cm

^{–1}, and the Y-sensitivity is 5 mV cm^{–1}.
What are the period and the peak
positive voltage of waveform in the diagram?

period
/ ms peak positive voltage / mV

A 5
17

B 5
25

C 10
17

D 10
25

**Reference:**

*Past Exam Paper – June 2007 Paper 1 Q4*

__Solution 542:__**Answer: C.**

The period is obtained by
considering the waveform horizontally.

Timebase setting is 2.5 ms cm

^{–1}.
Period is represented by 4cm =
4(2.5) = 10ms

The peak positive voltage may be
obtained by considering the maximum value vertically above the horizontal line.

Y-sensitivity is 5 mV cm

^{–1}.
Peak positive value is represented
by 3.5cm = 3.4(5) = 17mV

__Question 543: [Electric fields]__
Small water droplet of mass 3.0 μg
carries a charge of –6.0 × 10

^{–11}C. Droplet is situated in the Earth’s gravitational field between two horizontal metal plates. Potential of the upper plate is +500 V and potential of the lower plate is –500 V.
What is the motion of the droplet?

A It accelerates downwards.

B It remains stationary.

C It accelerates upwards.

D It moves upwards at a constant
velocity.

**Reference:**

*Past Exam Paper – June 2013 Paper 11 Q13*

__Solution 543:__**Answer: C.**

Electric field strength, E = V / d =
(500 – -500) / (2×10

^{-3}) = 5.0×10^{5}Vm^{-1}
Since the small water droplet is
negatively charged, it will experience an upwards force towards the positive
plate.

Electric force, F = Eq = (5.0×10

^{5}) (6.0×10^{–11}) = 3.0×10^{-5}N
The gravitational force of the Earth
causes a downward force on the water droplet due to its weight.

Weight = (3.0×10

^{-9}) (9.81) = 2.94×10^{-8}N

__Question 544: [Radian]__**(a)**Define the radian.

**(b)**A telescope gives clear view of distant object when angular displacement between edges of object is at least 9.7 × 10

^{−6}rad.

(i) Moon is approximately 3.8 × 10

^{5}km from Earth. Estimate minimum diameter of a circular crater on Moon’s surface that can be seen using telescope
(ii) Suggest why craters of same
diameter as that calculated in (i) but on surface of Mars are not visible using
this telescope

**Reference:**

*Past Exam Paper – June 2014 Paper 42 Q7*

__Solution 544:__**(a)**The radian is defined as the angle subtended at the centre of a circle by an arc equal in length to the radius.

**(b)**

(i)

{Since the angular displacement is
very small (9.7 × 10

^{−6}rad), the distance from the edge of the moon to the telescope (r) can be approximated to be equal to the distance of the moon from the telescope on Moon (3.8 × 10^{5}km).
180° correspond to π (which is about 3.14) rad. You can
convert the angular displacement given into degrees to have a better idea how
small it is.

Length of arc = rθ

Similarly, the length of the arc
formed can be approximated to be equal to the diameter of the moon since the
angular displacement is too small.

That’s why the question says ‘estimate’.}

Length of arc = distance x angle

Minimum diameter of circular crater
= (3.8x10

^{5}) (9.7x10^{-6}) = 3.7km
(ii) Mars is (much) further from
Earth / away. So, the angle (at the telescope is much) smaller.

Hey! Can you please explain O/N 2012 p43 Q6 b(ii), and c. Thanks!! :)

ReplyDeleteIt's explained as solution 546 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-107.html

Can you please explain Summer 2012 Paper 23 question 6(b)?

ReplyDeleteCheck solution 927 at

Deletehttp://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-190.html

In Q 543 shouldnt we first convert mass in grams to kg before finding weight

ReplyDeleteYou are right. Thanks. It has been corrected.

Delete