CSULA
I.
MIDTERM SOLUTION
Winter 2007
Professor: Rafael Obregon
PHYSICS 212
CONCEPTUAL QUESTIONS
1.
2.
d)
2T
e) 2v
m
k
T1  2
and T2 
2m
m
 2
 2T1.
k
k
By increasing the mass of the hanging object by a factor of two, you have increased the
tension by a factor of two. Since the rubber band has doubled its length, its mass per unit
length or linear mass density has gone down by a factor of two. Therefore,
T
v1 

and
v2 
2Te
1
2


4Te

2
Te

 2v1.
3.
Sketch the shape of the string (use a solid line).
4.
c) Bends away from the normal.
Because the light is entering a material, in which the index of refraction is lower, the speed of the light is
higher and the light bends away from the normal. By Snell’s Law:
sin 1 v2 n2
 
sin  2 v1 n1
5.
If: n2  n1  v2  v1
Therefore:  2  1
LEAD appears inverted but the OXIDE does not. Explain.
Both words are inverted. However, OXIDE has up-down symmetry whereas LEAD does not.
II) Solve
1.
Show that the force constant (k) of the spring has an expression of k = 16m.
T  2
2.
m
m
4 2 m 4 2 m 4  2 m
 T 2  4 2  k 


 16m
2
k
k
T2
2
 
 
4
2
A sinusoidal wave on a string has a wave function: y  x, t   sin  x  t  . If it’s mass per unit length is
10g/cm, find:
A) Comparing y  x, t   sin  x  t  with the wave function: y  A sin(kx  t )
Angular wave number:
B) Wavelength
k
k 1
rad
rad
rad
 1 2  100
cm
10 m
m
2
2
6.28rad
 

 0.0628m

k 100rad / m
C) Comparing y  x, t   sin  x  t  with the wave function: y  A sin(kx  t )
Angular frequency:   1
D) Wave speed
v

k

rad
s
1rad / s
 0.01m / s
100rad / m
3
E) Tension in the string: v  T  T  v 2   102 m / s 2 10 g 10 kg / g   104 N


2

 cm 10 m / cm 
F) Transverse speed of any element of the string
v   A cos(kx  t )  (1)(1) cos(1x  1t )   cos( x  t )
G) Transverse acceleration of any element of the string
a   2 A sin(kx  t )  (1)2 (1)(sin(1x  1t )   sin( x  t )
3.
Two pulses traveling in the same string are described by:
A) At what time do the two pulses cancel everywhere?
To cancel:
y1  y2  0
4
 2 x  3t  12 2  5

or
y1   y2 
4
 2 x  3t 2  5
For the positive root: 6t  12 
  2 x  3t    2 x  3t  12   2 x  3t    2 x  3t  12 
2
t2s
2
(at t  2 s , the waves cancel everywhere).
B) At what point do the two pulses always cancel?
For the negative root: 4 x  12  x  3 m
4.
(at x  3 m , the waves cancel always).
A concave mirror has a focal length of 30.0 cm.
For a concave mirror, R and f are positive.
For an upright image, M is positive. Therefore: M  
q
 3  q  3 p .
p
1 1 1
1
1 1
2
 

 p  20cm .
  
30.0 cm p 3 p 3 p
f
p q
5.
The light beam showed in the figure makes an angle of 20.0° with the normal line NN’ in the linseed oil.
A) Find the value of θ
Applying Snell’s law at the air-oil interface: nair sin   noil sin 20.0    30.4 .
B) Find the value of θ’
Applying Snell’s law at the oil-water interface: nw sin    noil sin 20.0 
   22.3 .