Sound Waves: Characteristics and Applications — CBSE Class 9 Science Important Questions

Answer: (c) wavelength.

One compression together with the following rarefaction makes one complete wave, so the distance between two consecutive compressions equals one wavelength $\lambda$.

Answer: (d) vacuum.

Sound is a mechanical wave and needs a material medium (solid, liquid or gas) whose particles can vibrate. A vacuum has no particles, so sound cannot propagate through it.

Answer: (b) $1.33\text{ m}$1.33 m.

Using $v=f\lambda$v=f, $\lambda=\dfrac{v}{f}=\dfrac{340}{256}\approx1.33\text{ m}$=v/f=340/2561.33 m.

Answer: (b) Frequency stays $500\text{ Hz}$500 Hz and wavelength increases.

Frequency is fixed by the source, so it does not change with the medium. Since $\lambda=\dfrac{v}{f}$=v/f and $v$v increases in water, $\lambda$ increases: from $\dfrac{340}{500}=0.68\text{ m}$340/500=0.68 m to $\dfrac{1500}{500}=3\text{ m}$1500/500=3 m.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Because the ear retains a sound for $0.1\text{ s}$0.1 s, the reflected sound must return after at least this interval to be heard separately. The sound covers a round trip $2d$2d in $0.1\text{ s}$0.1 s: $2d=344\times0.1$2d=344×0.1, giving $d\approx17.2\text{ m}$d17.2 m.

Amplitude is the maximum displacement of a particle of the medium from its mean (rest) position. SI unit: metre (m).

Time period is the time taken to complete one full oscillation (or for one wavelength to pass a point). SI unit: second (s).

Frequency $=100\text{ Hz}$=100 Hz means $100$100 vibrations per second.

In one minute $=60\text{ s}$=60 s: number of vibrations $=100\times60=6000$=100×60=6000.

Time period $T=\dfrac{1}{f}=\dfrac{1}{100}=0.01\text{ s}$T=1/f=1/100=0.01 s.

When a body such as a tuning fork vibrates, its forward motion pushes the neighbouring air particles together, creating a region of high pressure and density called a compression. Its backward motion leaves the particles farther apart, creating a low-pressure, low-density region called a rarefaction.

As the fork keeps vibrating, compressions and rarefactions are produced one after another and travel outward through the air, carrying energy. The air particles themselves only oscillate about fixed positions; they do not move along with the wave.

Sound is a longitudinal wave because the particles of the medium vibrate parallel (back and forth along) to the direction in which the wave travels.

The pulse travels down and the echo travels back, so it covers twice the depth in $3\text{ s}$3 s.

Total distance $=v\times t=1500\times3=4500\text{ m}$=v× t=1500×3=4500 m.

Depth $d=\dfrac{\text{total distance}}{2}=\dfrac{4500}{2}=2250\text{ m}$d=total distance/2=4500/2=2250 m.

This technique of finding depth/distance by timing reflected ultrasound is called SONAR (Sound Navigation And Ranging).

For each cliff the sound travels to the cliff and back, so distance to a cliff $=\dfrac{v\times t}{2}$=v× t/2.

Distance to first cliff $d_1=\dfrac{340\times1.5}{2}=255\text{ m}$d_1=340×1.5/2=255 m.

Distance to second cliff $d_2=\dfrac{340\times2.5}{2}=425\text{ m}$d_2=340×2.5/2=425 m.

Since the person stands between the cliffs, total distance $=d_1+d_2=255+425=680\text{ m}$=d_1+d_2=255+425=680 m.

(a)
Wavelength ($\lambda$): the distance between two consecutive points in the same phase (e.g. two successive compressions).
Frequency ($f$f): the number of complete waves passing a point per second.
Wave speed ($v$v): the distance travelled by the wave per second.

In one time period $T$T the wave advances by exactly one wavelength $\lambda$. Therefore
$v=\frac{\text{distance}}{\text{time}}=\frac{\lambda}{T}.$v=distance/time=/T.
Since $f=\dfrac{1}{T}$f=1/T, this gives $v=f\lambda.$v=f.

(b) $f=\dfrac{v}{\lambda}=\dfrac{330}{0.5}=660\text{ Hz}$f=v/=330/0.5=660 Hz.

Time period $T=\dfrac{1}{f}=\dfrac{1}{660}\approx1.52\times10^{-3}\text{ s}$T=1/f=1/6601.52×10^-3 s.

(a) An echo is the sound heard again after it is reflected from a distant surface and returns to the listener.

Because the sensation of sound persists in the ear for about $0.1\text{ s}$0.1 s, the reflected sound must reach the ear at least $0.1\text{ s}$0.1 s after the original for a distinct echo. In this time the sound makes a round trip $2d$2d:
$2d=v\times t=344\times0.1=34.4\text{ m}\ \Rightarrow\ d=17.2\text{ m}.$2d=v× t=344×0.1=34.4 m d=17.2 m.
So the reflector must be at least $17.2\text{ m}$17.2 m away.

(b) Applications of reflection of sound: (i) SONAR for measuring sea depth and locating objects under water; (ii) ultrasound scanning and stethoscopes / megaphones and soundboards in halls that use multiple reflections to direct sound.

Reverberation is the repeated reflection of sound from the walls, ceiling and floor of a hall that makes the sound persist and become muddled. It is reduced by covering the walls and ceiling with sound-absorbing materials such as curtains, carpets, fibreboard and perforated panels, and by using cushioned seats.

(i) The audible range is $20\text{ Hz}$20 Hz to $20000\text{ Hz}$20000 Hz ($20\text{ kHz}$20 kHz).

(ii) Ultrasound has a very short wavelength, so it travels in a straight, well-defined beam and reflects sharply from small cracks or defects instead of bending around them. This makes flaw detection accurate.

(iii) A bat (dolphins are also acceptable).

(iv) $\lambda=\dfrac{v}{f}=\dfrac{340}{50000}=6.8\times10^{-3}\text{ m}=6.8\text{ mm}$=v/f=340/50000=6.8×10^-3 m=6.8 mm.