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o-na [289]
3 years ago
5

Explain Sound level intensity with mathematical steps?

Physics
2 answers:
Sever21 [200]3 years ago
4 0

Answer:

In a quiet forest, you can sometimes hear a single leaf fall to the ground. After settling into bed, you may hear your blood pulsing through your ears. But when a passing motorist has his stereo turned up, you cannot even hear what the person next to you in your car is saying. We are all very familiar with the loudness of sounds and aware that they are related to how energetically the source is vibrating. In cartoons depicting a screaming person (or an animal making a loud noise), the cartoonist often shows an open mouth with a vibrating uvula, the hanging tissue at the back of the mouth, to suggest a loud sound coming from the throat Figure 2. High noise exposure is hazardous to hearing, and it is common for musicians to have hearing losses that are sufficiently severe that they interfere with the musicians’ abilities to perform. The relevant physical quantity is sound intensity, a concept that is valid for all sounds whether or not they are in the audible range.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is I=PAI=PA, where P is the power through an area A. The SI unit for I is W/m2. The intensity of a sound wave is related to its amplitude squared by the following relationship:

\(\displaystyle{I}=\frac{\left(\Delta{p}\right)^2}{2\rho{v}_{\text{w}}}\\\).

Here Δp is the pressure variation or pressure amplitude (half the difference between the maximum and minimum pressure in the sound wave) in units of pascals (Pa) or N/m2. (We are using a lower case p for pressure to distinguish it from power, denoted by P above.) The energy (as kinetic energy mv22mv22) of an oscillating element of air due to a traveling sound wave is proportional to its amplitude squared. In this equation, ρ is the density of the material in which the sound wave travels, in units of kg/m3, and vw is the speed of sound in the medium, in units of m/s. The pressure variation is proportional to the amplitude of the oscillation, and so I varies as (Δp)2 (Figure 2). This relationship is consistent with the fact that the sound wave is produced by some vibration; the greater its pressure amplitude, the more the air is compressed in the sound it creates

Sound intensity levels are quoted in decibels (dB) much more often than sound intensities in watts per meter squared. Decibels are the unit of choice in the scientific literature as well as in the popular media. The reasons for this choice of units are related to how we perceive sounds. How our ears perceive sound can be more accurately described by the logarithm of the intensity rather than directly to the intensity. The sound intensity level β in decibels of a sound having an intensity I in watts per meter squared is defined to be β(dB)=10log10(II0)β(dB)=10log10⁡(II0), where I0 = 10−12 W/m2 is a reference intensity. In particular, I0 is the lowest or threshold intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz. Sound intensity level is not the same as intensity. Because β is defined in terms of a ratio, it is a unitless quantity telling you the level of the sound relative to a fixed standard (10−12 W/m2, in this case). The units of decibels (dB) are used to indicate this ratio is multiplied by 10 in its definition. The bel, upon which the decibel is based, is named for Alexander Graham Bell, the inventor of the telephone.

Table 1. Sound Intensity Levels and IntensitiesSound .

chubhunter [2.5K]3 years ago
3 0

Answer:

sound intensity is explained by the following formula I= P/A where I= sound intensity(W/m²),P=power(W),A= area(m²) I hope this helps good luck!

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This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

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If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

Finally, the amplitude is:

x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

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