Trace the path of a ray emitted from the tip of the object through the focal point of the mirror and then the reflected ray that
results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray. Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded.
1 answer:
Answer:
1 / f = 1 / p + 1 / q
Explanation:
Mirror problems can be solved analytically with the constructor equation
1 / f = 1 / p + 1 / q
An approximate solution can also be given using graphical methods, where two of three rays must be drawn,
1) A ray that passes through the focal length and bounces horizontally
2) A ray that passes through the center of curvature (c = 2f) of the mirror and does not deviate
3) A beam that comes horizontal and bounces through the focal
The attached shows the ray that passes through the focal length and bounces horizontally.
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Answer:
T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N
Explanation:
a.)
Period: It is already given in the question "oscillator repeats its motion every 0.372 s".
So T=0.372 s
b)
frequency= f = 1/ T
f = 1/ 0.372
f=2.7 Hz
c).
Angular frequency= w= 2πf
w= 2*π*2.7
w=16.9 rad/s
d)
Spring Constant:
As w=
⇒w²= k/m
⇒k= m*w²
⇒k= 0.628 * 16.9² N/m
⇒k=179.2 N/m
e)
The mass will have maximum speed when it passes through the mean position.
At mean position
Maximum elastic potential energy = Maximum kinetic energy
1/2 k A² = 1/2 m v² ( A is amplitude of oscillation)
⇒ v=
⇒ v= \
⇒ v= 8.78 m/s
f)
Maximum force will be exerted on the block when it is at maximum distance.
F= k* A ( A is amplitude of oscillation)
F= 179.2 * 0.27 N
F= 48.4 N