HW #8 Solutions

10.There is atmospheric pressure outside the tire, so we find the net force from the gauge pressure. Because the reaction to the force from the pressure on the four footprints of the tires supports the automobile, we have

4PA = mg;

4(240103 N/m2 )(20010–4 m2) = m(9.80 m/s2), which gives m = 2.0103 kg.

29.When the balloon and cargo float, the net force is zero, so we have

Fnet = 0 = Fbuoy – mHeg – mballoong – mcargog;

0 = airgVballoon – HegVballoon – mballoong – mcargog;

0 = (1.29 kg/m3 – 0.179 kg/m3))p(7.35 m)3 – (1000 kg + mcargo),

which gives mcargo = 8.5102 kg.

6.The general expression for the displacement is x = A cos (t + , so

x(t = 0) = A cos .

(a)– A = A cos , which gives cos  = – 1, so  = p (or –p).

(b)0 = A cos , which gives cos  = 0, so  = p/2 (or 3p/2).

(c)A = A cos , which gives cos  = + 1, so  = 0.

(d)½ A = A cos , which gives cos  = ½ , so  = p/3 (or –p/3).

(e)– A /2 = A cos , which gives cos  = – !, so  = 2p/3 (or 4p/3).

(f)A /v2 = A cos , which gives cos  = 1/v2, so  = p/4 (or –p/4).

8.(a)We find the effective spring constant from the frequency:

f1 = (k/m1)1/2/2p;

2.5 Hz = [k/(0.050 kg)]1/2/2p, which gives k = 12 N/m.

(b)Because the size and shape are the same, the spring constant, which is determined by the

buoyant force, will be the same. The new frequency of vibration will be

f2 = (k/m2)1/2/2p = [(12 N/m)/(0.25 kg)]1/2/2p = 1.1 Hz.

30.Immediately after the collision, the block-bullet system will have its maximum velocity at the equilibrium position. We find this velocity from energy conservation:

Ki + Ui = Kf + Uf;

½ (M + m)v02 + 0 = 0 + ½ kA2;

½ (0.300 kg + 0.0125 kg)v02 = ½ (2.25103 N/m)(0.124 m)2, which gives v0 = 10.5 m/s.

We find the initial speed of the bullet from momentum conservation for the impact:

mv + 0 = (M + m)v0;

(0.0125 kg)v = (0.300 kg + 0.0125 kg)(10.5 m/s), which gives v = 263 m/s.

9.(a)Because the pulse travels up and back, the speed is

v = 2L/t = 2(600 m)/(16 s) = 75 m/s.

(b)The mass density of the cable is

 = m/L = AL/L = A.

We find the tension from

v = (FT/)1/2 = (FT/A)1/2;

75 m/s = [FT/(7.8103 kg/m3)p(0.7510–2 m)2]1/2 , which gives FT = 7.8103 N.

20.The traveling wave is

D = (0.48 m) sin [(5.6 m–1)x + (84 s–1)t].

(a)We find the wavelength from the coefficient of x:

(5.6 m–1)x = 2px/, which gives  = 1.12 m.

(b)We find the frequency from the coefficient of t:

(84 s–1)t = 2pft, which gives f = 13 Hz.

(c)From the positive sign between the x and t terms, the wave is traveling in the – x direction, with

speed

v = f = (13.4 Hz)(1.12 m) = 15 m/s (toward negative x).

(d)The amplitude is the coefficient of the sine function:

DM = 0.48 m.

(e)The speed of a particle is

u = ?D/?t = (0.48 m)(84 s–1) cos [(5.6 m–1)x + (84 s–1)t] = (40 m/s) cos [(5.6 m–1)x + (84 s–1)t].

Thus we have

umax = (40 m/s)(1) = 40 m/s;

umin = (40 m/s)(0) = 0.

3.The speed in the concrete is determined by the elastic modulus:

vconcrete = (E/)1/2 = [(20109 N/m2)/(2.3103 kg/m3)]1/2 = 2.95103 m/s.

For the time interval we have

Δt = (d/vair) – (d/vconcrete);

1.4 s = d{[1/(343 m/s)] – [1/(2.95103 m/s)]}, which gives d = 5.4102 m.

28.The wavelength of the fundamental frequency for a string is  = 2L, so the speed of a wave on the string is

v = f = 2(0.32 m)(196 Hz) = 125 m/s.

We find the tension from

v = [FT/(m/L)]1/2;

125 m/s = {FT/[(0.6810–3 kg)/(0.32 m)]}1/2 , which gives FT = 33 N.

62.Because the wavelength in front of a moving source decreases, the wavelength from the approaching source is

1 = (v – v1)/f0.

This wavelength approaches the stationary listener at a relative speed of v, so the frequency heard by the listener is

f1 = v/1 = v/[(v – v1)/f0]= vf0/(v – v1) = (343 m/s)(2000 Hz)/(343 m/s – 15 m/s) = 2091 Hz.

The wavelength from a stationary source is

2 = v/f0.

This wavelength approaches the moving receiver at a relative speed of v + v1, so the frequency heard is

f2 = (v + v1)/2 = (v + v1)/(v/f0) = (v + v1)f0/v = (343 m/s + 15 m/s)(2000 Hz)/(343 m/s) = 2087 Hz.

The two frequencies are not exactly the same, but close.

For the other speeds we have,

at 150 m/s:

f3 = v/1 = v/[(v – v1)/f0]= vf0/(v – v1) = (343 m/s)(2000 Hz)/(343 m/s – 150 m/s) = 3554 Hz;

f4 = (v + v1)/2 = (v + v1)/(v/f0) = (v + v1)f0/v = (343 m/s + 150 m/s)(2000 Hz)/(343 m/s) = 2875 Hz.

at 300 m/s:

f3 = v/1 = v/[(v – v1)/f0]= vf0/(v – v1) = (343 m/s)(2000 Hz)/(343 m/s – 300 m/s) = 15950 Hz;

f4 = (v + v1)/2 = (v + v1)/(v/f0) = (v + v1)f0/v = (343 m/s + 300 m/s)(2000 Hz)/(343 m/s) = 3750 Hz.

The Doppler formulas are not symmetric; the speed of the source creates a greater shift than the speed of the observer.

We can write the expression for the frequency from an approaching source as

f1 = v/1 = vf0/(v – v1) = f0/[1 – (v1/v)].

If v1«v, we use 1/[1 – (v1/v)] ˜ 1 + (v1/v), so we have

f1 ˜ f0[1 + (v1/v)],

which is the expression f2 = (v + v1)f0/v for the frequency for a stationary source and a moving listener.

Note that we have carried more significant figures than justified to show differences.