Physics 101 - Homework # 10 Solutions

• Problem 15-11
• Problem 15-16
• Problem 15-26
• Problem 15-28
• Problem 15-51
• Problem 16-10
• Problem 16-18
• Problem 16-19
• Problem 16-28

Chapter 15

Problem 11)

a) The frequencies and speeds are the same:

f = /2 (22 s^-1)/2

= 11 Hz

v = /k = (22 s^-1/(3 m^-1)

= 7.3 m/s

b) We use the trigonometric identity for the sum of two sin functions (just like we did in lecture) to get

z_l = 3sin(kx + t) and z_r = 3sin(kx - t)

Problem 16)

The pulse frequency is

f_pulse = |f₂-f₁|

2 Hz = |f₂-260 Hz|

which gives

f₂ = 262 Hz    or 266 Hz

Problem 26)

The wavelength of the sound is

= v/f = (330 m/s)/(600 Hz) = 0.55 m

from the analysis of Figure 15-11 in the text, we know that the path-length difference from the two sources separated by d to the point an angle from the centerline is

L = dsin

a) For the first minimum, this path-length difference must be ½:

sim₁ = L/d = ½/d = ½(0.55 m)/(1.4 m) = 0.196

which gives

₁ = 11.3°

The distance from the centerline is

D₁ = Ltan₁ = (10.0 m)tan 11.3°

= 2.0 m

Problem 28)

We know that the path-length difference from two sources, separated by d, to the point an angle from the centerline is

L = dsin

The distance from the centerline is

D = Ltan

For the first minimum L = /2. With D = 0.4 mm and L = 25 cm, we note that the angle is small, so we have, using the small angle approximation,

sin= tan

so

/2d = D/L

/2(0.2 x 10^-3 m) = (0.4 x 10^-3)/(25 x 10^-2 m)

which gives

= 6.4 x 10^-7 m

Problem 51)

The wavelength of the standing waves set up on the two violins are the same, so the difference in frequency implies a difference in wave velocity. Assuming the violins strings on the two violins are the same (i.e. same mass per unit length µ), the velocity difference implies a difference in their tensions. For the dependence of the frequency on the tension, we have

f = v/ = (1/)(T/µ)^1/2

The frequency change and thus the tension change will be small, so we treat them as differentials. Because the wavelengths and mass densities are the same, we have

f = ½(1/)(1/Tµ) ^1/2T

When we divide this by the first equation for the frequency, we get

f/f = ½T/T

(2 Hz)/(440 Hz) = ½T/T

which gives

T/T = 9.1 x 10^-3

or a little less than 1% different.

Chapter 16

Problem 10)

Keeping house windows slightly open will allow the pressure change to occur on both sides of the window and wall, so that there is no large unbalanced force on the wall.

F_net = pA = [(14.7 - 14.0 lb/in²](300 ft²) (12 in/ft)²

= 3.0 x 10⁴ lb    (1.3 x 10⁵ N)

Problem 18)

The pressure at a depth h is

P = P₀ + gh

Because p_gauge is P-P₀, we have

P_gauge = gh

(40 lb/in²)(4.45 N/lb)/(2.54 x 10^-2 m/in)² = (1.03 x 10³ kg/m³)(9.8 m/s²)h

which gives

h = 27 m

Problem 19)

The force on the large piston must equal the weight of the car. The pressures will be the same, so we have:

P = F/A₂ = mg/A₁

F/[(1/4)D₁²] = mg//[(1/4)D₂²]

or

F = mg(D₁/D₂)² =(1200 kg)(9.8 m/s²)[(30 cm)/(2 cm)]²

= 52 N

With no frictional losses, the work done by F must increase the potential energy of the car:

Fh₁ = mgh₂

(52 N)(0.5 m) = (1200 kg)(9.8 m/s²h₂

which gives

h₂ = 2.2 x 10^-3 m

= 2.2 mm

Note that this can also be obtained by equating the changes in volume.

Problem 28)

When the balloon floats, the net force is zero, so we have (remember that the helium is not massless!)

F_buoy = m_Heg + Mg

_airgV_balloon = _HegV_balloon + Mg

which can be rearranged to

(_air - _H)(4/3)R³ = M

(1.3 kg/m³- 0.18kg/m³)(4/3)R³ = 100 kg

Solving for R gives

R = 2.8 m