course phys 202

-The longitudal waves had a higher velocity, but the amplitude of the wave affects intensity.

2.Compare the intensity of sound at 120 dB with that of a whisper at 20 dB.

-dB = 10 log(I / I-threshold)

The equation than needs to be altered to solve for I (I = I_threshold * 10^12), then after using algebra the final equation is = 10 log(I_1 / I_threshold) - ( 10 log(I_2 / I_threshold). After substutiing we get 120-20=100, =10 log(I_1 / I_2) and

3.length of open pipe, 262 Hz at 21 C? **** gen phy What is the length of the pipe?

- Using the equation v = (331 + 0.60 * Temp.), we substitute 21C, and get v=343.6 m/s. Then to find the wavelength we substitute the velocity and frequency and get .33 meters. Wavelength = 1.3 m. Since its an open pipe there are antinodes at both ends, the wavelength is twice the length so it is 1.3m/2= .64m

Good work.

course phys 202

-the power of a 20.5cm lens is 1/ .205m= 4.8m^-1 dp this indicates a converging lens because it has a positive focal length

A lens with -6.25 dp is 1/-6.25m^-1= -.16m, and this is diverging because it has a negative focal length

2.incident at 45 deg to equilateral prism, n = 1.52; and what angle does light emerge?

-One should use snells law to figure the angle of refraction at the surface (n1sin'thea1=n2sin'thea2)

n1= 1.00, n2=1.52

1.00sin45 degrees=1.52sin'theta2

'thea 2=27.7 degrees

Angle of incidence: (90-'thea2)+(90-'thea3)+angle at top of

triangle=180degrees.

62.3 degrees+(90-'thea3)+60 degrees=180 degrees-'thea)=57.7degrees…

thea=32.3

Since these angles are parallel, one can calculate the refraction of the second surface,

nsin'thea3=n(air)sin''thea4

1.52sin32.3=1.00sin (thea4)

'thea 4=54.3 degrees

&#Let me know if you have questions. &#