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Given the following displacement – time graph for a mass oscillating on a spring. i. period ii. Find the frequency of the above wave 14. Two girls stood 200m from a wall. One of the girls banged two p

Physics

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2 parts
i

2.0 s\text{2.0 s}

ii

0.5 Hz\text{0.5 Hz}

333.33 m/s\text{333.33 m/s}

1.92×10−4 m3/s\text{1.92} \times 10^{-4}\text{ m}^3\text{/s}

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  1. Given the following displacement – time graph for a mass oscillating on a spring.

Part i· 1 step

period

  1. Identify one complete oscillation from the graph.

    A complete oscillation is the time taken for the wave to repeat its pattern. From the graph, the wave starts at zero displacement, goes up, down, and returns to zero displacement, moving in the same direction, at 2.0 s2.0\text{ s}.
    Alternatively, it reaches a peak at 1.0 s1.0\text{ s} and the next peak is at 3.0 s3.0\text{ s}. The time between two consecutive peaks is the period.
    Period T=2.0 sT = 2.0\text{ s}.
    The period is 2.0 s\boxed{\text{2.0 s}}.

Part ii

0.5 Hz\text{0.5 Hz}

Find the frequency of the above wave

  1. Use the relationship between frequency and period.

    f=1Tf = \frac{1}{T}

    Where ff is frequency and TT is period.

  2. Substitute the period found in part (i).

    f=12.0 sf = \frac{1}{2.0\text{ s}} f=0.5 Hzf = 0.5\text{ Hz}

    The frequency of the wave is 0.5 Hz\boxed{\text{0.5 Hz}}.

    1. Two girls stood 200m200\text{m} from a wall. One of the girls banged two pieces of wood together while the second started a stop watch and stopped it when she heard the echo. If the time shown on the stop watch was 1.2 sec1.2\text{ sec} determine the speed of the sound.
  3. Determine the total distance traveled by the sound.

    The sound travels from the girls to the wall and then reflects back to the girls (echo).
    Distance to wall = 200 m200\text{ m}
    Total distance traveled by sound, d=2×200 m=400 md = 2 \times 200\text{ m} = 400\text{ m}

  4. Use the formula for speed.

    Speed=DistanceTime\text{Speed} = \frac{\text{Distance}}{\text{Time}}

    Given time, t=1.2 st = 1.2\text{ s}

  5. Substitute the values and calculate the speed of sound.

    Speed=400 m1.2 s\text{Speed} = \frac{400\text{ m}}{1.2\text{ s}} Speed=333.33 m/s\text{Speed} = 333.33\text{ m/s}

    The speed of the sound is 333.33 m/s\boxed{\text{333.33 m/s}}.

    1. A lawn sprinkler has 40 holes each of cross section area 2.0×10−2 cm22.0 \times 10^{-2}\text{ cm}^2. The sprinkler is connected to a hose pipe of cross section area 1.6 cm21.6\text{ cm}^2, if the speed of the water in the hose pipe is 1.2 m/sec1.2\text{ m/sec}, calculate the flow rate in the hose pipe
  6. Identify the given values for the hose pipe.

    Cross-section area of hose pipe, Ahose=1.6 cm2A_{\text{hose}} = 1.6\text{ cm}^2
    Speed of water in hose pipe, vhose=1.2 m/sv_{\text{hose}} = 1.2\text{ m/s}

  7. Convert the area of the hose pipe to square meters.

    Ahose=1.6 cm2=1.6×(10−2 m)2=1.6×10−4 m2A_{\text{hose}} = 1.6\text{ cm}^2 = 1.6 \times (10^{-2}\text{ m})^2 = 1.6 \times 10^{-4}\text{ m}^2
  8. Calculate the flow rate in the hose pipe using the formula Q=A×vQ = A \times v.

    Qhose=Ahose×vhoseQ_{\text{hose}} = A_{\text{hose}} \times v_{\text{hose}} Qhose=(1.6×10−4 m2)×(1.2 m/s)Q_{\text{hose}} = (1.6 \times 10^{-4}\text{ m}^2) \times (1.2\text{ m/s}) Qhose=1.92×10−4 m3/sQ_{\text{hose}} = 1.92 \times 10^{-4}\text{ m}^3\text{/s}

    The flow rate in the hose pipe is 1.92×10−4 m3/s\boxed{\text{1.92} \times 10^{-4}\text{ m}^3\text{/s}}.
    (Note: The information about the sprinkler holes is not needed to calculate the flow rate in the hose pipe.)

    1. State two factors affecting boiling point of a liquid
    • External pressure: A higher external pressure increases the boiling point, while a lower external pressure decreases it.
    • Impurities: The presence of dissolved impurities (solutes) in a liquid generally increases its boiling point.
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