Formative Year 12

1
Formative Year 12

• Physics Examination.

2
In a class experiment to find the speed of sound
a starting pistol was fired by the teacher while
some girls measured the time between seeing the
smoke and hearing the bang at the end of the oval.
The timings were then repeated with the teacher
and the girls exchanging their positions, still
maintaining the 240 m separation. The class
results are shown
3
From these results the most accurate value for
the speed of sound on the oval that day would be
about A. 339 ms-1 B. 341 ms-1 C. 343 ms-1 D.
346 ms-1

• Working The Average time for the experiment was
  0.69333 s.
• This was found by simply adding up all the times
  and dividing by the
• number of data points, in this case, it was 6.
• ?using velocity displacement ? time
• V 240 ? 0.693
• V 346.1538 m/s

4
In the last question the teacher pointed out that
the speed of sound from East to West was
different to that from West to East because of a
wind blowing on the day. From the results above
an approximate value for the wind speed is
about A. 2.7 ms-1 B. 3.3 ms-1 C. 3.7 ms-1 D.
4.1 ms-1
Working For East to West The Average time for
the experiment was 0.686667s. ?using velocity
displacement ? time ? V 240 ? 0.686667 ? V
349.5146 m/s Then the other way West to East The
Average time for the experiment was 0.7s. ?using
velocity displacement ? time ? V 240 ? 0.7 ?
V 342.8571 m/s The difference between the two
values is 6.657 m/s Now the difference is twice
the speed of the wind. In other words, the speed
of the wind was 3.33m/s as in one direction, the
speed of the wind would be added to the speed of
sound, but subtracted when going the opposite
way.
5

• The speed of sound in a certain gas is related to
  the temperature of the gas by the formula v
  405 0.085T
• The calculated speed of sound in this gas at a
  temperature of 25 oC would be about
• 407 ms-1
• 409 ms-1
• 411 ms-1
• 413 ms-1

• Working
• ?V 405 0.085 ? 25
• ?v 407.125 ms-1

6

• An ultrasonic sound wave is sent from one end of
  a 150 m long pipe filled with the same gas
  referred to in the previous question at a
  temperature of 51oC so it hits the other end of
  the pipe and rebounds back to the transmitter

• The time taken under these conditions, for the
  wave to travel to the end of the pipe and back
  again is about
• A. 0.37 s
• B. 0.73 s
• C. 0.81 s
• D. 0.86 s

• Working The speed of sound in
• that gas is 409.33 ms-1. Using
• Velocity displacement ? time
• Time displacement ? velocity
• Time 300 ? 409.33
• Time 0.73s

7
A yacht sends an ultrasonic sonar pulse from a
transmitter on its keel downwards towards the sea
bed. The outgoing pulse (X) and the reflected
pulse (Y) are shown on a CRO screen, as displayed
on the screen. The speed of sound in seawater
was 1520 ms-1.

• Working The difference between
• X and Y is 4 divisions or 80 ms.
• ? Using v s ? t
• ? s v ? t
• s 121.6m
• But this the total distanced travelled.
• So the actual depth is half that
• distance, which is 60.8m.

If the time scale on the screen is 1 horizontal
division 20 milliseconds then the depth of the
water from the bottom of the boat was about A.
25 m B. 46 m C. 61 m D. 120 m
8
A siren is made by blowing air at a high pressure
through hole in a rotating steel disc.

• The disc shown is of diameter of 62 cm, has 200
  holes in its outer edge and rotates at a rate of
  1500 revolutions per minute. The frequency of the
  sound emitted from the siren would be about
• 5 kHz
• 8 kHz
• 10 kHz
• 90 kHz

Working The disc makes 1500 revolutions in one
minute, hence it will make 25 revolutions in 1
second. With 200 holes in each revolution there
will be 25 ? 200 5000 holes per
second. Therefore the frequency of the sound will
be 5000 Hertz or 5 kHz.
9

• A firework explodes in the sky releasing 20 W of
  sound energy in a time of 0.05 s. John is
  standing 80 m away at the time when he hears the
  bang. The intensity of sound at Johns ear will
  be about
• A. 1.5 x 10-3 Wm-2
• B. 2.5 x 10-3 Wm-2
• C. 5 x 10-3 Wm-2
• D. 2.0 x 10-2 Wm-2

• Working
• Now it is assumed that sound radiates in a
  spherical fashion with a surface area of 4?r2.
• ?I P? (4?r2)
• ?I 20 ? (4??802)
• ?I 2.5?10-4 Wm-2
• But it was only for 0.05s
• So the intensity was
• ?I 2.5?10-4 ? 0.05
• ?I 5 ? 10-3 Wm-2

10

• At the firework display mentioned in Q. 7, Jim is
  standing 20 m away when the same firework
  explodes. The ratio of sound intensities heard by
  Jim and John respectively is
• A. 4 1 B. 8 1
• C. 16 1 D. 24 1

• Working As he is actually 4 times closer than
  John, the Intensity if inversely proportional to
  the square of the distance. Therefore 4 times
  closer yields a factor of 42 or 16 times greater
  intensity.

11
A fast speed boat moving at 5 ms-1 send a pulse
of ultrasonic waves downwards towards the bed of
a river so that they are received back at a later
time ahead of its present position.

• Working
• v s ? t
• ? t s ? v
• t 50 ? 1500
• t 0.033s
• or 33 ms.

• The speed of sound in water is 1500 ms-1 and the
  riverbed is 25 m below the boats hull when the
  ultrasonic pulse was sent. The approximate time
  for the beam transmitted beam to be received
  again by the boat is
• 1 ms
• B. 12 ms
• C. 25 ms
• D. 33 ms

12

• The velocity of a wave down a guitar string is
  given by the formula
• where
• T tension in the string
• m mass per metre of the string in kgm-1
• The top E string on a guitar has a total mass of
  0.208 g and is stretched to a tension of 226 N.
  The total length of the string is 62.8 cm.
• The velocity of the wave down this string would
  be about
• 346 ms-1
• 576 ms-1
• 633 ms-1
• 826 ms-1

13

• The diagram shows the wave pattern for resonance
  in
• a closed tube when a speaker emitting a frequency
  f is
• placed above.
• The wave starts a small distance e above the
  tube.
• The formula for the velocity of the wave down the
  tube will be
• v 4f(L e)
• v (L e) ? 4f
• v 4Lf e
• v (4f e) ? L

• Working This is a vibrating column of air with
  one end closed
• Therefore, the L n/4 l where n harmonic
  number, in this case n1
• l 4L.
• To include the end correction
• l 4(Le).
• But v f l.
• ? v f ? 4(Le).

14

• The loudspeaker in the previous question has a
  very low frequency signal fed into it that is
  gradually increased until resonance occurs. The
  frequency is increased several times more,
  obtaining several other values of resonant
  frequencies. If the speed of sound is 360 ms-1, L
  is 90 cm and e is 2 cm, the frequencies where
  resonance occurs will be around
• A. 98 Hz, 196 Hz, 392 Hz
• B. 98 Hz, 293 Hz, 490 Hz
• C. 96 Hz, 192 Hz, 384 Hz
• D. 96 Hz, 288 Hz, 480 Hz

• Working
• v f ? 4(Le).
• 360 f ? 4(0.90.02)
• ? f 360 ? 3.68
• f 97.8 Hz
• Hence the fundamental frequency of this setup is
  98 Hz.
• For higher harmonics,
• Fn nF1 where F1 fundamental
• ? Fn 3 ? 98 294 Hz
• ? Fn 5 ? 98 490 Hz

15
A guitar string is 75 cm long between the bridge
and the nut, shown vibrating in its fundamental
mode.

• If the speed of a transverse wave in this string
  is 850 ms-1 at what frequency will the string
  vibrate in its third harmonic?
• 566 Hz B. 840 Hz
• C. 1160 Hz D. 1700 Hz

Working L n/2 l ?l 2L?n ?l 2 ? 0.75 ? 1 ?l
1.5 m The fundamental wavelength 1.5m Using v
f l. ?f v ? l ?f 850 ? 1.5 566.7 Hz So
the third harmonic, Fn nf1 ?f 3 ? 566.7
1700.1 Hz
16
Which wave pattern represents the highest
pitched, loudest note?
Answer B Logic Look for the waveform that has
the highest frequency meaning that the wavelength
is the shortest. Also, the amplitude determines
the loudness, hence B also has the largest
amplitude.
17

• Cathode ray Oscilloscope Trace CRO) Horizontal
  Setting is 20 ?s per cm In the picture opposite
  the time-base setting on the CRO is 20 ?s per
  division. A single note is fed into the CRO from
  a synthesizer that produces the trace shown. The
  frequency of the note is about
• 83 Hz
• 270 Hz
• 1.9 kHz
• 8.3 kHz

Working Assuming each division is 1cm, therefore
the Length of 1 wavelength is 6cm or 120?s. This
is the period of the Note. The period is the
reciprocal of frequency. Hence, the frequency is
8333.3 Hz or 8.3 kHz.
18
Which of the following classification of waves is
correct?
Longitudinal Transverse A Water waves,
sound Electromagnetic, X-rays B Seismic P
waves, ultrasonic Violin string, ripples C Flute
waves, red light ß waves, blue light D Seismic
S waves, ? waves Guitar string, UV waves
Answer The answer is B
19
A tight rubber cord is held in a girls hand,
attached to a wall 4 m away and pulled tight. It
is known that the speed of a pulse in the cord at
this tension is 2.5 ms-1.

• Answer The frequency of waves
• is 5 Hz. With a velocity of 2.5ms-1,
• ? l 2.5 ? 5 0.5m
• The waves will meet at a distance
• of 0.25m from the wall, which is half
• the time difference between 2
• successive pulses or 0.2s.
• But it takes 1.6s for the first pulse to
• get to the wall and 1.8s for the second
• pulse to get to the wall. But the
• difference between pulses is 0.2s, so it
• must take 1.7s before the pulses start
• interfering. ie a 1.60.1s.

The girl produces 5 pulses per second traveling
down the cord that rebound back from the wall as
shown. What time, after the pulses are started
will the first reflected pulse traveling back
meet the second pulse traveling forward along the
cord? A. 1.7 s B. 1.9 s C. 2.2 s D. 2.5 s
20
A tuning fork of frequency f is struck and held
above a plastic tube half-submerged in water, as
shown. The length of tube out of the water (L) is
adjusted until a fundamental resonance position
is obtained. The experiment is repeated for four
other tuning forks of different frequencies and
four other resonant lengths are obtained.

• Which graphical plot (y versus x) would yield a
  straight line of best fit, assuming no end
  correction applies?
• f versus L
• L versus f2
• L2 versus f
• f versus 1/L

• Working since L n/4 l.
• But v f l.
• l v ? f
• Hence L (n/4) v ? f (n/4) v ? f-1
• Hence a graph of L Vs f-1 would yield a graph
• With a straight line of (n/4)v.

21

• A tuning fork with an unknown frequency is
  sounded at the same time as a guitar string of
  vibrating length 140 cm and 5 beats per second
  were produced. When the vibrating length of the
  guitar string was altered to 142 cm and sounded
  with the tuning fork a different beats frequency
  of 3 per second was heard. It is also known that
  the frequency of a guitar string is inversely
  proportional to its length.
• From these results we can deduce that the
  frequency of the tuning fork was
• 130 Hz
• 137 Hz
• 141 Hz
• 143 Hz
• (Hint set up simultaneous equations)

• Guitar 1
• fbeat fguitar ffork
• ?5 v ? 2 ? 1.4 ffork (1)
• Guitar 2
• fbeat fguitar ffork
• ?3 v ? 2 ? 1.42 ffork (2)
• Solving Simultaneously by first
• Rearranging (1) to get
• ffork v ? 2.8 5 (1a)
• And substituting (1a) into (2) to get
• 3 v ? (2.84) v ? (2.8) 5
• Solving for v, you get v 397.6 ms-1
• Substituting v into (2), you get
• 3 397.6 ? (2.84) - ffork
• Hence ffork is therefore 137 Hz.