(week 6) wave and sound

Wave and sound

(1-0) types of wave

Water wave

Sound waves travel in air with a speed of 343m/s. The lowest frequancy sound we can hear is 20.0 Hz; the highest frequency is 20.0kHz. Find the wave length of sound for frequencies of 20.0Hz and 20.0kHz

Solution

Solve Equation for λ

λ = v/f = (343 m/s)/(20.0s⁻¹) = 17.2 m

λ = v/f = (343 m/s)/(20,000s⁻¹) = 1.72 m

(2.0) wave and string

The speed of a wave on a string

A 5.0-m length of rope, with a mass of 520g, is pulled taut with a tension of 46N. Find the speed of waves on the rope.

Solution

First, calculate the mass per length, μ:

520g = 0.52kg

μ = m/ L = (0.52kg)/(5.0m) = 0.10kg/m

Now, substitute μ and F into Equation (2.0):

ν = √(F/ μ) = √[46N/(0.10kg/m)]

(3.0) harmonic wave functions

A loudspeaker puts out 0.15 W of sound through a square area 2.0 m on each side. What is the intensity of this sound?

Solution

Applying Equation 14-5, with A = (2.0m)², we find

I = intensity = ?

P = power = 0.15W

A = area = (2.0m)²

I = P/A = (0.15W)/(2.0m)² = 0.038W/m²

(4.0) harmonic wave functions

The power of song

Two people relaxing on a deck listen to a songbird sing. One person, only 1.00m from the bird, hears the sound with an intensity of 2.80x10⁻⁶W/m². (a) what intensity is heard by the second person , who is 4.25m from the bird? Assume that no reflected sound is heard by either person. (b) What is the power output of the bird’s song?

Our problem, the two observers, one at a distance of r1 = 1.00m from the bird, the other at a distance of r2 = 4.25m. The sound emitted by the bird is assumed to spread out spherically, with no reflections

Strategy

a. the two intensities are related by equation 4.0 ,with r1 = 1.00m and r2 = 4.25m.

b. the power output can be obtained from the definition of intensity, I =( P/A). we can calculate P for either observer, noting that A = 4πr²

solution

part (a)

1. Substitute numerical values into equation (4.0)

I2 = (r1/r2)²I1 = (1.00m/4.25m)²(2.80 X 10⁻⁶ W/m²)

= 1.55 X 10⁻⁷ W/m²

Part (b)

2. Solve I =( P/A) for the power ,p, using data for observer 1:

I1 =( P/A1)

P = I1 A1 = (2.80 X 10⁻⁶W/m²)[4π(1.00m)²]

= 3.52 X 10⁻⁵W

3. As a check, report the calculation for observer 2:

I2 =( P/A2)

P = I2A2 = (1.55 X 10⁻⁷W/m²)[4π(4.25m)²]

= 3.52 X 10⁻⁵W

(5.0)

At a distance of 5m from a source the sound level is 90dB. How far away has the level dropped to 50dB

Solution

I1 = (P/4πr1²) and I2 = (P/4πr2²), so (I2 /I1) = (r1²/r2²)

Β1 = 10 log (I1 /I0 ) = 90dB , so (I1 /I0 ) = 10⁹

similarly (I2 /I0 ) = 10⁵

thus (I2 /I1) = (10⁵/10⁹) = 10⁻⁴ = (r1²/r2²), so r2 = 10²r1 = 500m