MCQ 11 Mark
What is known as strength of ear's perception of sound?
- ✓Loudness
- BIntensity
- CFrequency
- DAmplitude
Answer
View full question & answer→Correct option: A.
Loudness
Loudness is the strength of ear's perception of sound.
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This is because the screams are ultrasonic $($above $20,000$ hertz$)$ which the human ear cannot hear as humans can only hear the sound in the range of $20\ Hz − 20,000\ Hz.$
$30,000Hz$
Sound having a frequency of more than $20000Hz$ in known as ultrasonic. It can not heard by us.
The speed of water in sound is $1530m/s.$
The sound which is produced due to a mixture of several frequencies is called a note and it is pleasant to listen too. An octave consists of eight different note, ranging from $256Hz$ to $512Hz.$
In solids, there are two kinds of acoustic waves -- longitudinal [the medium is vibrating in the direction of the acoustic wave] and transverse [the medium is vibrating at right angles to the direction of propagation]. The speed of sound is greater for longitudinal waves. The trend in the speed of sound is: the harder the material and the lower density - the faster is the speed of sound.
The approximate value for the speed of sound in air is $344\ m/ s.$
Distance $= 300\ m$
Speed $= 5000\ m/ s$
$\therefore$ $\text{Time taken}=\frac{\text{Distance}}{\text{Time}}$
$=\frac{300}{5000}$
$=0.06\text{s}$
In longitudinal wave motion, the particles in the medium travel parallel to the direction of propagation of wave.
In the given case, since the loaded spring applies force only in axial direction of the spring, hence the movement is also axial so the waves produced are longitudinal waves.
$SONAR$ uses ultrasonic sound waves.
given,
height, $s = 500m$
$g = 10$
Speed of sound $= 340$
On analysing we get that we have to find time period
first analyze the data and apply different formulas for it then see which formula suits the best. The formula that should be used here is
$\text{s}=\text{ut}+\frac{1}{2\text{at}^2}$
$\text{u}=0$
$500+1+\frac{1}{2}\times10\times\text{t}\times\text{t}$
We get $t = 10\ sec$
Now the time for the resound $=\frac{500}{340}=1.47\text{sec}$
Total time $= 10 + 1.47$
$= 11.47\ sec$
The frequency of a wave is not altered by crossing a boundary. The reflected pulse becomes inverted when a wave in a less dense rope is heading towards a boundary with a more dense rope. The amplitude of the incident pulse is always greater than the amplitude of the reflected pulse.
The frequency $(f)$ of a wave is the number of full wave forms generated per second. This is the same as the number of repetitions per second or the number of oscillations per second.
In this case, an object attached to one end of a spring makes $20$ vibrations in $10 s.$ That is, $20$ vibration in $10$ seconds. S in one second it makes $2$ vibrations.
Time taken to list the sent signal is $0.4s.$
Speed of sound in water is $1,500\ m/s.$
So,
$=\frac{1500\times0.4}{2}$
$=300\text{m}$
Sound travels faster in liquids than in gases because molecules are more tightly packed. In fresh water, sound waves travel at $1,482$ meters per second $($about $3,315\ mph).$ That's well over 4 times faster than in air. In fact, sound waves travel over $17$ times faster through steel than through air.
Transmitter produces and transmits ultrasonic waves of very high frequency and detector converts the reflected ultrasonic waves into electric signals.
The ultrasound waves can penetrate into matter to a large extent because they have very high frequency. They have frequency greater than $20,000\ Hz.$
Some of the elephant's vocalizations are infrasonic, and thus are inaudible to humans. The prominence of very low frequencies in the vocalizations is a defining characteristic of all three species of elephants. Sounds are generally considered to be infrasonic if their frequency is less than $20Hz ($the lower limit of human hearing$).$ Low frequency sounds travel farther than high frequency ones, which make them ideal for long distance communication.