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Figure 1 shows a bat and its prey - Edexcel - GCSE Physics - Question 1 - 2023 - Paper 1

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Figure 1 shows a bat and its prey. The bat emits a high frequency sound pulse to locate its prey. The speed of sound in air is 330 m/s. (a) The wavelength of the s... show full transcript

Worked Solution & Example Answer:Figure 1 shows a bat and its prey - Edexcel - GCSE Physics - Question 1 - 2023 - Paper 1

Step 1

Calculate the frequency of the sound.

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Answer

To find the frequency of the sound, we will use the formula:

v=f×λv = f × λ

Where:

  • v is the speed of sound (330 m/s)
  • f is the frequency (in Hz)
  • λ is the wavelength (11 mm or 11×10−311 × 10^{-3} m)

Substituting the values, we rearrange the equation to find f:

f=vλ=33011×10−3f = \frac{v}{λ} = \frac{330}{11 × 10^{-3}}

Calculating this gives:

f=30000 Hzf = 30000 \text{ Hz}

Step 2

Calculate the distance from the bat to its prey.

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Answer

The total time for the sound pulse to travel to the prey and back is 18 ms (or 18×10−318 × 10^{-3} s). Therefore, the time taken to reach the prey is half of this time:

t=18×10−32=9×10−3 st = \frac{18 × 10^{-3}}{2} = 9 × 10^{-3} \text{ s}

Using the formula for distance, which is given by:

extdistance=v×t ext{distance} = v × t

Substituting the known values:

distance=330 m/s×9×10−3 s\text{distance} = 330 \text{ m/s} × 9 × 10^{-3} \text{ s}

Calculating this gives:

distance=2.97 m\text{distance} = 2.97 \text{ m}

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