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Complete the sentences - AQA - GCSE Physics - Question 7 - 2022 - Paper 1

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Complete the sentences. Choose the answers from the box. The distance between the centre of one compression of a sound wave and the centre of the next compression i... show full transcript

Worked Solution & Example Answer:Complete the sentences - AQA - GCSE Physics - Question 7 - 2022 - Paper 1

Step 1

Complete the sentences. Choose the answers from the box.

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Answer

The distance between the centre of one compression of a sound wave and the centre of the next compression is called the wavelength. The number of waves passing a point each second is called the frequency.

Step 2

Complete the sentence. Choose the answer from the box.

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Answer

In a longitudinal wave, the oscillations are parallel to the direction of energy transfer.

Step 3

A sound wave has a frequency of 8.0 kHz. Which of the following is the same as 8.0 kHz? Tick (✓) one box.

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Answer

The equivalent frequency to 8.0 kHz is 8000 Hz.

Step 4

Calculate the period of a sound wave with a frequency of 8.0 kHz. Use the Physics Equations Sheet.

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Answer

To find the period of a sound wave, we use the formula:

Period(T)=1frequency\text{Period} (T) = \frac{1}{\text{frequency}}

Plugging in the frequency:

T=18000=0.000125 sT = \frac{1}{8000} = 0.000125 \text{ s}

Step 5

Calculate the wavelength of a sound wave with a frequency of 6600 Hz. speed of sound = 330 m/s

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Answer

Using the formula for wavelength:

wavelength(λ)=speedfrequency\text{wavelength} (\lambda) = \frac{\text{speed}}{\text{frequency}}

Substituting the values:

λ=3306600=0.05 m\lambda = \frac{330}{6600} = 0.05 \text{ m}

Step 6

Write down the equation which links distance (s), speed (v) and time (t).

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Answer

The equation that links distance, speed, and time is:

s=vts = vt

Step 7

Distance A on Figure 14 is 13.2 m. speed of sound = 330 m/s. Calculate the time taken for the sound to travel from loudspeaker A to the technician.

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Answer

Using the rearranged equation:

t=dvt = \frac{d}{v}

Substituting the known values:

t=13.23300.04 st = \frac{13.2}{330} \approx 0.04 \text{ s}

Step 8

Explain why the sound from loudspeaker B should be emitted slightly before the sound from loudspeaker A.

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Answer

Loudspeaker B is further from the technician than speaker A, so the sound would take more time to travel (to the technician) and thus, arrives at the technician at the same time.

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