(i) Velocity of both the waves P and Q is the same, since it depends on the medium through which they pass which is the same in both the cases.
| (iii) (Frequency of P) / (Frequency of Q) | = | fP / fQ |
| = | [ v/λP] / [v/λQ] | |
| or fP / fQ | = | λQ / λP (since v is same for both) |
| = | 3 / 2 | |
| i.e., pitch of P is higher than that of Q. |
(iv) The wave Q has a larger amplitude than P. Therefore Q will produce a louder sound than P.
The human ear can detect continuous sounds in the frequency range from 20 Hz to 20000 Hz. Assuming that the speed of sound in air is 330 m s –1 for all frequencies, calculate the wave lengths corresponding to the given extreme frequencies of the range ?
| From relation v | = | f λ |
| For frequency f | = | 20 Hz, |
| ∴ λ | = | v / f |
| = | 330 / 20 | |
| = | 16.5 m | |
| For frequency f | = | 20000 Hz, |
| ∴ λ | = | 330 / 20000 |
| = | 16.5 × 10–3 m | |
| = | 16.5 mm. |
A certain sound has a frequency of 280 hertz and a wavelength of 1.2 m. Calculate the speed with which this sound travels ? What difference would be felt by a listener between this sound and another sound traveling at the same speed but of wavelength 2.8 m ?
| For first sound, f | = | 280 Hz, |
| λ | = | 1.2 m |
| Speed v | = | f × λ |
| = | 280 × 1.2 | |
| = | 336 ms–1 | |
| For second sound, v | = | 336 ms–1, |
| λ | = | 2.8 m |
| Frequency, f | = | v / λ |
| = | 336 / 2.8 | |
| = | 120 |
| Given, wave length of red light | = | 5000 Å, |
| velocity of light | = | 3 × 108 ms–1 |
| (i) Frequency | = | ? |
| ∴ v | = | f × λ |
| 3 × 108 ms–1 | = | f × 5000 × 10–10 |
| ∴ f | = | 3 × 108 / 5000 × 10–10 |
| = | 6 × 1014 Hz | |
| (ii) Time | = | ? |
| ∴ Time (T) | = | 1 / Frequency |
| T | = | 1 / f |
| = | 1 / 6 × 1014 | |
| = | 1.67 × 10–15 s |
| Given frequency | = | 450 Hz, |
| velocity | = | 336 ms–1 |
| (i) Wave length (λ) | = | ? |
| ∴ v | = | f × λ |
| 336 | = | 450 × λ |
| λ | = | 336 / 450 |
| = | 0.747 m | |
| (ii) Time (T) | = | ? |
| ∴Time (T) | = | 1 / Frequency |
| T | = | 1 / f |
| = | 1 / 450 | |
| = | 0.0022 s |
| Case (i) | ||
| Wave length of a sound wave λ | = | 18 cm |
| = | 0.18 m | |
| Frequency of a sound wave | = | 1500 Hz |
| Since, v | = | f × λ |
| = | 0.18 × 1500 | |
| = | 270 ms–1 | |
| Case (ii) | ||
| λ | = | 22 cm |
| = | 0.22 m | |
| ∴ v | = | f × λ |
| 270 | = | f × 0.22 |
| ∴ f | = | 270 / 0.22 |
| = | 1227.27 Hz |
| Distance between one crest and trough | = | λ / 2 |
| ∴ λ / 2 | = | 8 cm |
| λ | = | 16 cm |
| Frequency of vibrator | = | 2800 / min |
| = | 2800 / 60 s | |
| = | 46.7 Hz | |
| (i) Time period | = | 1 / f |
| = | 1 / 46.7 | |
| = | 0.0214 s | |
| (ii) Wave velocity | = | f × λ |
| = | 46.7 × 16 | |
| = | 747.2 cm s–1 | |
| = | 7.472 ms–1 |
If 18 waves are produced per second, what is the frequency in hertz ?
The frequency in hertz is equal to the number of waves produced per second. In this case, since 18 waves are being produced per second, so the frequency of the waves is 18 hertz (which is also written as 18 Hz).
The wavelength of sound emitted by a source is 1.4 × 10–2m. Calculate frequency of the sound, if its velocity is 354.2 ms–1 ?
The relationship between velocity, frequency and wavelength of a wave is given by the formula
| v | = | f × λ |
| Here, velocity, v | = | 354.2 ms–1 |
| Frequency, f | = | ? (To be calculated) |
| And, wavelength, λ | = | 1.4 × 10–2m |
| So, putting these values in the above formula, we get : | ||
| 354.2 | = | f × 1.4 × 10–2 |
| f | = | 354.2 / (1.4 × 10–2) |
| = | (3542 × 102 )/ 14 | |
| = | 253 × 102 Hz | |
| Thus, the frequency of sound is 253 × 102 hertz. |
(a) Frequency
We know that frequency of a wave is the number of waves produced in 1 second.
| Here, No. of waves produced in 3 seconds | = | 18 |
| So, No. of waves produced in 1 second | = | 18 / 3 |
| = | 6 |
So, the frequency of this wave is 6 hertz.
(b) Wavelength
| so, 12 cm | = | λ / 2 |
| And, λ | = | 12 × 2 cm |
| = | 24 cm |
Thus, the wavelength of this wave is 24 centimeters.
| Now, v | = | f × λ |
| So, | = | 6 × 0.24 m/s |
| = | 1.44 m/s |
Thus, the velocity of this wave is 1.44 metre per second.
In this case we have to find out the period of the waves. But to do that, we have to first find out the frequency of the waves from the given data. Please note that the distance between two consecutive crests is equal to the wavelength of the wave. so,in this case the wavelength will be 125m.
| Now, velocity of wave, v | = | 20 m/s |
| Frequency of wave, f | = | ? (To be calculated) |
| And, Wavelength of wave, λ | = | 125 m |
| We know that : v | = | f × λ |
| So, 20 | = | f × 125 |
| f | = | 20 / 125 |
| = | 0.16 Hz | |
| Now, period of a wave is given by the relation : | ||
| T | = | 1 / f |
| So, | = | 1 / 0.16 |
| (or) | = | 6.25 s |
i.e, the crests reach the boat after every 6.25 seconds.
Sound waves travel with a speed of about 330 m/s. What is the wavelength of sound whose frequency is 660 hertz ?
| Here, speed of waves, v | = | 330 m/s |
| Frequency of waves, f | = | 660 Hz |
| And, wavelength, λ | = | ? (To be calculated) |
| Now, v | = | f × λ |
| So, 330 | = | 660 × λ |
| λ | = | 330 / 660 |
| = | 0.5 m | |
| Thus, the wavelength of sound waves is 0.5 metre. |