CSULA
I.
1.
MIDTERM
PHYSICS 212
Since: a   2 x , then:  2  k1  k 2
m
F
net
→
Professor: Rafael Obregon
  k1 x    k 2 x   ma  a  
k1  k 2
x.
m
k1  k 2
m
Two harmonic waves are described by y1  (5 m) sin3x  6t  & y 2  (5 m) sin3x  6t  . Write an expression for the resultant standing
wave y = y1 + y2. What is the magnitude of the speed (in m/s) of the two traveling waves?
y = y1 + y2 = (10m) sin (3x) cos (6t)
3.
Fall- 2012
CONCEPTUAL QUESTIONS (10 points ea)
The mass in the figure slides on a frictionless surface. Use Hooke’s Law to write the equation of motion of the system,
then find an expression for the angular frequency ω in function of k1 , k2 and m.
The equation of motion then becomes:
2.
SOLUTION
v   / k  6rad / s  / 3rad / m  2m / s
&
The laws of refraction and reflection are the same for sound as for light. The speed of sound in air is 340 m/s and in water
it is 1510 m/s. If a sound wave approaches a plane water surface at an angle of incidence of 10, what is the angle of
refraction?
From Snell’s Law: sin  1  v A   2  sin 1  sin 10  1510   50 .5 
sin  2
4.

vW
340 
A diverging lens has radius of curvature of 20.0 cm. Find the image’s size and position of an object placed at the focal
length of the lens. (Sketch may help but it is not necessary)
For a diverging lens: f = R/2 is negative → f = -10cm.
So: p = 10cm. → 1  1  1 →
p q
f
1
1
1
1
1
1
2
 
 


 q  5cm
10 cm q  10 cm
q
10 cm 10 cm 10 cm
M 
q 5cm

 0.5
p 10 cm
II) Open-Ended Problems (12 points ea)
1. A simple harmonic oscillator of amplitude A has a total mechanical energy Emec. Determine:
(A). The potential energy (in function of Emec) when the position is one-third the amplitude.
If : x = (⅓)A → U s  12 k 13 A2  19 12 kA2  19 Emech


(B). The kinetic energy (in function of Emec) when the position is one-third the amplitude.
Emech  U s  K  Emech 
1
9
Emech  K  K 
8
9
Emech
(C). For what values of the position (in function of A) does the kinetic energy equal one-half the potential energy?
Emech  U s  K  12 kA2 
1
2
kx2  12

1
2

kx2  A2  x2  12 x2  x   2 / 3 A
2. A block of mass M, supported by a string, rests on a frictionless incline making an angle θ with the horizontal (See the figure).
The length of the string is L, and its mass is m << M.
(A). Taking the positive x-direction to be up the incline and using Newton’s 2nd law, derive an expression for the tension T in the
string.
F
x
sin0


TMg (A free-body diagram may help but it is not necessary)


TMg
sin
or the tension in the string is:

(B). Derive an expression for the speed of a transverse wave in the string in function of M, m, L, g and θ. (Recall μ is the mass per
unit
/M
sinsin
m
v
T



gMgL
Lmlength)
(C). Derive an expression for the time interval (Δt) required for a transverse wave to travel from one end of the string to the other,
in
function of M, m, L , g and θ.

vMgLMg
tL
LmmL
sinsin

3. The A string on a cello vibrates in its first normal mode with a frequency of 220 Hz. The vibrating segment is 70.0 cm long and
has a mass of 1.20 g.
(A). Find the wave length of the first normal mode.
Wavelength for a String of Length L: n = 2L/n (n = 1) → 1 = 2L = 1.40m
(B). Find the tension in the string.
f  v 
T

 (1.40 )( 220 )m / s 
T
1.2  10 /(0.70)  T  163 N
3
(C). Determine the frequency of vibration when the string vibratesf=
segments
3220
3
³in three
Hz =660
Hz (loops).
With one-third the distance between nodes, the frequency is
4.
Assume a transparent rod of diameter d has an index of refraction of n = 1.36 (see figure).
(A).
Find a value for the critical angle θc (total internal reflection) inside the transparent rod.
n
1.00
sin   air 
 0.735 → c  47.3
c
npipe
1.36
(B). Using Geometry, find a value for angle Φ (refracted angle for the first end of the rod).   90.0  c  90.0  47.3  42.7
(C). Determine the maximum angle θ for which the light rays incident on the end of the rod are subject to total internal reflection
along the walls of the rod.
Snell’s law at the end, 1.00sin   1.36sin 42.7 gives   67.2
5.
A convex mirror has a radius of curvature of 40.0 cm. Determine the object position for which the resulting image is virtual,
upright and one-half times the size of the object.
For a convex mirror: f = R/2 is negative → f = -20cm. Also: M   q  h'   12 h  1  q   1 p →
p
h
h
1 1
1 →1
1
1
1 2
1
1
1
 


  


 p  20 cm
p q
f
p  1 p  20 cm
p p
20 cm
p
20 cm
2
2
2