AQA · GCSE · Physics

    Course version: 8463 · Version from Sept 2016

    The time taken for the reflections to reach a detector can be used to…

    Full specification wording

    Waves for detection and exploration (physics only) (HT only): The time taken for the reflections to reach a detector can be used to determine how far away such a boundary is.

    The measurement is a timing.

    Checked against the AQA 8463 specification (opens in a new tab) on .

    • Point 4 of 12
    • 4 key terms

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    The measurement is a timing. A short pulse is sent out, travels to the boundary, reflects and travels back, and the detector records the time t between transmission and echo. In that time the pulse has covered twice the depth, so the distance to the boundary is (v × t) / 2, where v is the speed of the wave in that medium. Forgetting the factor of two is the classic error because it doubles the answer. For a boundary 6 cm deep in tissue where the speed is 1500 m/s, the echo returns after 2 × 0.06 / 1500 = 8 × 10⁻⁵ s. If two echoes are recorded, the separation between the two boundaries comes from the difference in their times, using the same halving rule.

    What earns the marks

    • Using speed = distance / time with the speed in the correct medium.
    • Halving the calculated distance because the pulse travels to the boundary and back.
    • A correct substitution with the time converted to seconds.
    • The final depth given with a correct unit.

    Common mistakes and how to put them right

    • Leaving out the factor of two, so quoting a depth twice the true value.

    • Using the speed of sound in air for a pulse travelling through tissue or water.

    • Leaving the time in milliseconds or microseconds while the speed is in m/s.

    • Taking the gap between two echoes as the time when the question asks for the depth of the first boundary.

    Exam tips

    • Write total distance = 2 × depth as your first line so the halving cannot be forgotten.
    • Convert µs and ms to seconds before substituting, and show the converted value.
    • Check the size of your answer against the context; a scan depth of 30 m should tell you something is wrong.

    Try the lesson

    The start of the Waves for detection and exploration (physics only) (HT only): The time taken for the reflections to reach a detector can be used to determine how far away such a boundary is. lesson, made from checked teaching notes for this specification point.

    Have a go first

    A pulse of ultrasound is sent into a block of steel. A reflection returns to the detector 0.000 08 s after the pulse was emitted. The speed of ultrasound in steel is 5900 m/s. Calculate the distance from the detector to the boundary that produced the reflection.

    Listen

    The measurement is a timing. A short pulse is sent out, travels to the boundary, reflects and travels back, and the detector records the time t between transmission and echo. In that time the pulse has covered twice the depth, so the distance to the boundary is (v × t) / 2, where v is the speed of the wave in that medium. Forgetting the factor of two is the classic error because it doubles the answer. For a boundary 6 cm deep in tissue where the speed is 1500 m/s, the echo returns after 2 × 0.06 / 1500 = 8 × 10⁻⁵ s. If two echoes are recorded, the separation between the two boundaries comes from the difference in their times, using the same halving rule. A strong answer earns marks for these points. using speed = distance / time with the speed in the correct medium. halving the calculated distance because the pulse travels to the boundary and back. a correct substitution with the time converted to seconds. the final depth given with a correct unit. Exam tip: Write total distance = 2 × depth as your first line so the halving cannot be forgotten.

    See it

    A top-down flowchart with eight nodes. The first node says a short pulse is sent out and points to the second, where the pulse travels to the boundary. That points to the third, where the pulse reflects at the boundary, which points to the fourth, where the echo travels back to the detector. The fourth points to the fifth, where the detector records the time t between transmission and echo. The fifth points to the sixth, giving the total distance travelled as v times t. The sixth points to the seventh, noting the pulse has covered twice the depth, and the seventh points to the final node, giving the distance to the boundary as v times t divided by two.
    The sequence from sending a pulse to calculating the distance to a boundary, showing why the total distance is halved.

    Put it in order

    Put these steps in the order they happen.

    1. 1. Detector records time t between transmission and echo
    2. 2. Distance to boundary = (v × t) / 2
    3. 3. Total distance travelled = v × t
    4. 4. Short pulse sent out
    5. 5. Pulse travels to the boundary
    6. 6. Pulse has covered twice the depth
    7. 7. Pulse reflects at the boundary
    8. 8. Echo travels back to detector

    Key words

    Say what each word means before you turn the card.

    echo timing

    Try answering aloud or in your head before turning the card.

    there and back distance

    Try answering aloud or in your head before turning the card.

    speed equals distance over time

    Try answering aloud or in your head before turning the card.

    unit conversion

    Try answering aloud or in your head before turning the card.

    See every definition
    echo timing
    A short pulse is sent out, travels to the boundary, reflects and travels back, and the detector records the time between transmission and echo.
    there and back distance
    In the recorded time the pulse has covered twice the depth, so the distance to the boundary is (v × t) / 2.
    speed equals distance over time
    Using speed = distance / time with the speed in the correct medium, then halving the calculated distance because the pulse travels to the boundary and back.
    unit conversion
    Converting microseconds and milliseconds to seconds before substituting, and showing the converted value.

    Label the diagram

    Some boxes have lost their names. Pick the right name for each number.

    The diagram again, with 3 boxes numbered instead of named. A top-down flowchart with eight nodes. The first node says a short pulse is sent out and points to the second, where the pulse travels to the boundary. That points to the third, where the pulse reflects at the boundary, which points to the fourth, where the echo travels back to the detector. The fourth points to the fifth, where the detector records the time t between transmission and echo. The fifth points to the sixth, giving the total distance travelled as v times t. The sixth points to the seventh, noting the pulse has covered twice the depth, and the seventh points to the final node, giving the distance to the boundary as v times t divided by two.

    • 1Box 1

    • 2Box 2

    • 3Box 3

    0 of 3 labelled

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