Physics 192 Solutions to Mastering Physics Week 4 Assignment

Physics 192 Solutions to Mastering Physics Week 4 Assignment  P15.1. Prepare: The wave is a traveling wave on a stretched string. We will use Equation 15.2 to find the wave speed. Solve: The wave speed on a stretched string with linear density  is vstring  TS /  . The wave speed if the tension is doubled will be

2TS

  vstring Assess:



 2 vstring  2 (200 m/s)  280 m/s

Wave speed increases with increasing tension.

P15.50. Prepare: The wave pulse is a traveling wave on a stretched string. We will use Equations 15.2 and 15.1 to find the string mass. Solve: The wave speed on a stretched string with linear density  is vstring  Assess:

TS





TS  m /L

LTS 2.0 m (2.0 m)(20 N)    m  0.025 kg  25 g 3 m 50  10 s m

A mass of 25 g for the 2.0 m long string is realistic.

  P15.14. Prepare: The wave is a traveling wave. A comparison of the given wave equation with Equation 15.7 yields A  3.5 cm, 2 /   2.7 rad/m, and 2 / T  92 rad/s. Solve: (a) The frequency is

f  which is 15 Hz to two significant figures. (b) The wavelength is 2



1  1  2 92 rad/s    14.6 Hz T  2  T 2

 2.7 rad/m   

2  2.33 m 2.7 rad/m

which will be reported as 2.3 m to two significant figures. (c) The wave speed is v   f  34 m/s.

  P15.18. Prepare: Please refer to Figure P15.18. Solve: The amplitude of the wave is the maximum displacement which is 6.0 cm. The period of the wave is 0.60 s, so the frequency f  1/ T  1/0.60 s  1.667 Hz, or 1.7 Hz to two significant figures. The wavelength, using Equation 15.10, is v 2 m/s    1.2 m f 1.667 Hz Assess:

It is important to know how to read information from graphs.

  P15.60. Prepare: Obtaining information from Figure P15.60 and using basic sound related information, we can determine the desired quantities. Solve: (a) The wavelength is read directly from the graph as   0.12 m. (b) Referring to Figure P15.60, it looks like the wave has traveled a distance of about  x  0.012 m in a time  t  0.01 s to give a wave speed of v   x /  t  0.012 m/0.01 s  1.2 m/s. (c) The frequency of the wave is f  v /   (1.2 m/s)/(0.12 m)  10 Hz. (d) The amplitude appears to be around 4mm. Assess: It is important to know how to obtain information from a pictorial representation.

 

P15.64. Prepare: We need to first find the wavelength and the period to write the y-equation for a wave traveling in the positive x-direction. We will determine the wavelength using Equation 15.10. The y-equation that we are asked to write will look like Equation 15.7. Solve: The wavelength and the period are calculated as follows:

v 400 m/s  2m f 200 Hz 1 1  0.005 s T  f 200 Hz



The displacement equation for the wave is   x  t  y ( x, t )  (0.010 mm)cos  2     2 m 0.005 s   Assess:

The negative sign in the cosine function’s argument indicates motion along the +x direction.

P15.66. Prepare: Knowing the speed and wavelength of the wave (periodic disturbance) moving along the string, we can determine the frequency of the disturbance. This frequency tells us the rate at which a point on the string is moving up and down (i.e., its linear frequency). Knowing the linear frequency, we can determine the angular frequency which can be combined with the amplitude information to determine the maximum translational speed of any point on the string. Solve: The frequency of the disturbance and hence the linear frequency of the oscillating particle on the string is determined by f  v /  (24 m/s)/(0.3 m)  80 Hz. The maximum transverse speed of any point on the string is determined by vmax   A  2 fA  2 (80 Hz)(102 m)  5.0 m/s. Assess: It is important to distinguish between the speed of the wave along the string and the transverse speed of a point on the string.

  P15.28. Prepare: Follow Example 15.12. The intensity at the stage is I stage  (1.0  1012 W/m 2 )10(110 dB/10 dB)  0.10 W/m 2

We do not know the power of the source unless we assume a distance for the given 110 dB data; so let’s assume 1.0 m for the distance from the speaker to the edge of the stage. Equation 15.13 shows that the intensity at 30 m will be (30 m) 2/(1 m) 2  900 times less if we assume that the wave is spherical. Therefore the new intensity is 0.10 W/m2  1.11  104 W/m 2 900 The power (or energy/time) is the intensity multiplied by the area. We can deduce from the information given that the area of the eardrum is a   R 2   (4.2 mm) 2  5.54  105 m 2. Solve: P  Ia  (1.11 104 W/m 2 )(5.54  105 m 2 )  6.2  109 W So the eardrum receives 6.2  109 J each second during the concert. Assess: This doesn’t seem like much energy, but prolonged exposure to loud sounds like this can damage your hearing.

  P15.71. Prepare: We will use Equation 15.12 and the definition of power. Solve:

(a) The peak power of the light pulse is

Ppeak  (b) The average power is

E 500 mJ 0.500 J    5.0  107 W t 10 ns 1.0  108 s

Pavg 

Etotal 10  500 mJ 5 J   5W 1s 1s 1s

The laser delivers pulses of very high power. But the laser spends most of its time “off,” so the average power is very much less than the peak power. (c) The intensity is I laser 

P 5.0  107 W 5.0  107 W    6.4  1017 W/m 2 2 a  (5.0  m) 7.85  1011 m 2

(d) The ratio is

I laser 6.4  1017 W/m 2   5.8  1014 I sun 1.1  103 W/m 2 Assess: The laser produces light whose intensity is more than 14 orders of magnitude larger than the sun’s intensity on the target.