Physics, 13.01.2021 15:20 sheilashair

A whistle emitting a note of 4 kHz is whirled in a horizontal circle of radius 1 m at a constant speed. If the highest note heard
by a person a large distance away is 4200 Hz, find:
(i) the speed of the whistle
(ii) the lowest note heard by the person,
(iii) the time taken for the whistle to make on complete revolution,
(iv) the time interval between the person hearing the highest and lowest note.

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Answers

Answer from: Alexandragurule18

(i) The speed of the whistle is 16. $\overline 3$ m/s

(ii) The lowest note heard by the person is 3,818. $\overline{18}$ Hz

(iii) The time taken for the whistle to make one revolution is approximately 0.38468 seconds

(iv) The time interval between the person hearing the highest and lowest note is approximately 0.19234 seconds

Explanation:

The question relates to the Doppler effect for which the change in frequency is given as follows;

$f_L = \left ( \dfrac{v \pm v_L }{v \pm v_S} \right ) \cdot f_S$

Where;

f_L = The frequency heard by the listener

v = The speed of sound in air = 343 m/s

v_L = The listener's speed

v_S = The source's (whistle) speed

f_S = The frequency of the source

The given parameters are;

The frequency of the whistle (source), f_S = 4 kHz = 4,000 Hz

The radius of the horizontal circle around which the whistle is whirled = 1 m

The highest note heard by a person a large distance away = 4200 Hz

The distant listener is taken as stationary, v_L = 0 m/s

We therefore have;

$f_L = \left ( \dfrac{v \pm 0 }{v \pm v_S} \right ) \cdot f_S = \left ( \dfrac{v }{v \pm v_S} \right ) \cdot f_S$

(i) When the whistle's motion while it is being whirled towards the listener, v_S is negative

When the whistle's motion while it is being whirled is away from listener, v_S is positive

The highest note, $f_{LH}$ , is obtained with the lowest denominator, that is when the source moves towards the listener and v_S is negative, we get;

$f_{LH} = 4,200 \ Hz = \left ( \dfrac{343 \ m/s }{343 \ m/s - v_S} \right ) \times 4,000 \ Hz$

$\dfrac{4,200 \ Hz}{4,000 \ Hz} = \dfrac{21}{20} = \dfrac{343 \ m/s}{343 \ m/s - v_S}$

21·(343 m/s - v_S ) = 20×343 m/s

21 × v_S = 21 × 343 m/s - 20 × 343 m/s = 343 m/s

v_S = (343 m/s)/21 = 49/3 m/s = 16. $\overline 3$ m/s

The speed of the source = The speed of the whistle

∴ The speed of the source = v_S = 16. $\overline 3$ m/s = The speed of the whistle

(ii) The lowest note heard by the person occurs when the denominator of the formula for the change in frequency is smallest for which we have the value of v_S as positive and the source moves away from the observer which is presented as follows;

$f_{LH} = \left ( \dfrac{343 \ m/s }{343 \ m/s +16.\overline 3 \ m/s } \right ) \times 4,000 \ Hz = 3,818.\overline {18} \ Hz$

The lowest note heard by the person = 3,818. $\overline{18}$ Hz

(iii) The angular velocity of the whistle, ω is given as follows;

$\omega = \dfrac{\theta}{t} = \dfrac{v_S}{r} = \dfrac{16. \overline 3 \ m/s}{1 \ m} = 16. \overline 3 \ radians/second$

The angle rotated in one revolution, is 2·π radians, therefore, when θ = 2·π radians, the time, t_c , to complete one revolution is given as follows;

$t_c = \dfrac{\theta}{\omega} = \dfrac{2 \times \pi \ radians}{16. \overline 3 \ radians/second} \approx 0.385 \ seconds$

Therefore, the time to complete one revolution ≈ 0.38468 seconds

(iv) The time interval between the person hearing the highest and lowest note is the time to complete half a cycle ≈ 0.38468/2 seconds ≈ 0.19234 seconds.

Answer from: Quest

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation:

Answer from: Quest

im sorry i dont know

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