Reflection of sound waves enables bats to hunt for moths. The sound wave produced by a bat has a frequency of 222 kHz and a wavelength of 1,5 x 10^{-3} m. 6.1 Calculate the speed of this sound wave through the air. 6.2 A stationary bat sends out a sound signal and receives the same signal reflected from a moving moth at a frequency of 230,3 kHz. 6.2.1 Is the moth moving TOWARDS or AWAY FROM the bat? 6.2.2 Calculate the magnitude of the velocity of the moth, assuming that the velocity is constant.

NSC Physical Sciences Doppler Effect: Reflection of sound waves enable

Reflection of sound waves enables bats to hunt for moths - NSC Physical Sciences - Question 6 - 2016 - Paper 1

Question 6

Reflection of sound waves enables bats to hunt for moths. The sound wave produced by a bat has a frequency of 222 kHz and a wavelength of 1,5 x 10^{-3} m. 6.1 Calcu... show full transcript

Worked Solution & Example Answer:Reflection of sound waves enables bats to hunt for moths - NSC Physical Sciences - Question 6 - 2016 - Paper 1

Step 1

Calculate the speed of this sound wave through the air.

96%

114 rated

Answer

To find the speed of the sound wave, we can use the formula:

$v = f \lambda$

where:

$v$ is the speed of the wave,
$f$ is the frequency (222 kHz),
$\lambda$ is the wavelength (1.5 x 10^{-3} m).

Substituting the values:

$v = (222 \times 10^3)(1.5 \times 10^{-3})$

Calculating this gives:

$v = 333 \text{ m/s}$

Step 2

Is the moth moving TOWARDS or AWAY FROM the bat?

99%

104 rated

Answer

The moth is moving TOWARDS the bat since the frequency of the reflected sound signal (230.3 kHz) is greater than the original frequency (222 kHz). This indicates that the source (moth) is approaching the observer (bat).

Step 3

Calculate the magnitude of the velocity of the moth, assuming that the velocity is constant.

96%

101 rated

Answer

We can use the Doppler effect formula to calculate the velocity of the moth:

$f_r = f_s \frac{v + v_o}{v - v_s}$

Given:

$f_r = 230.3 \text{ kHz}$ (frequency received),
$f_s = 222 \text{ kHz}$ (frequency sent),
$v = 333 \text{ m/s}$ (speed of sound),
$v_o = 0$ (bat is stationary).

Rearranging the formula:

$f_r (v - v_s) = f_s v$

Substituting the known values:

$230.3 \times 10^3 (333 - v_s) = 222 \times 10^3 (333)$

Simplifying this:

$76689.9 - 230.3 v_s = 73926$

Now solving for $v_s$ gives:\n $v_s = \frac{76689.9 - 73926}{230.3} \approx 12 \text{ m/s}$

Reflection of sound waves enables bats to hunt for moths - NSC Physical Sciences - Question 6 - 2016 - Paper 1

Worked Solution & Example Answer:Reflection of sound waves enables bats to hunt for moths - NSC Physical Sciences - Question 6 - 2016 - Paper 1

Calculate the speed of this sound wave through the air.

Is the moth moving TOWARDS or AWAY FROM the bat?

Calculate the magnitude of the velocity of the moth, assuming that the velocity is constant.

Join the NSC students using SimpleStudy...