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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

(a) Calculate the frequency of the sound.

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Answer

To calculate the frequency of the sound, we can use the equation:

v=f×λv = f \times \lambda

Where:

  • vv = speed of sound (330 m/s)
  • λ\lambda = wavelength of the sound (11 mm = 11 \times 10^{-3} m)

Rearranging the formula to solve for frequency (ff):

f=vλf = \frac{v}{\lambda}

Substituting the values:

f=33011×10−3f = \frac{330}{11 \times 10^{-3}}

Evaluating this gives:

f=30000 Hzf = 30000 \text{ Hz}

Step 2

(b) Calculate the distance from the bat to its prey.

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Answer

To calculate the distance from the bat to its prey, we can use the formula:

Distance=Speed×Time\text{Distance} = \text{Speed} \times \text{Time}

The speed of sound is 330 m/s and the time taken for the pulse to return is 18 ms (which is 18 \times 10^{-3} seconds).

Thus, the distance is:

Distance=330 m/s×18×10−3 s\text{Distance} = 330 \text{ m/s} \times 18 \times 10^{-3} \text{ s}

Calculating this gives:

Distance=5.94 m\text{Distance} = 5.94 \text{ m}

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