Monday, April 6, 2015

Physics 9702 Doubts | Help Page 106

  • 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) = 5.0×105 Vm-1

Since the small water droplet is negatively charged, it will experience an upwards force towards the positive plate.
Electric force, F = Eq = (5.0×105) (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

Since the downward weight is much smaller than the upward electric force, the droplet experiences a resultant upward force which causes it to accelerate upwards.






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 × 105 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 × 105 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 = (3.8x105) (9.7x10-6) = 3.7km

(ii) Mars is (much) further from Earth / away. So, the angle (at the telescope is much) smaller.



6 comments:

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

    ReplyDelete
    Replies
    1. It's explained as solution 546 at
      http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-107.html

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

    ReplyDelete
    Replies
    1. Check solution 927 at
      http://physics-ref.blogspot.com/2015/08/physics-9702-doubts-help-page-190.html

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

    ReplyDelete
    Replies
    1. You are right. Thanks. It has been corrected.

      Delete

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