The position of an 14.6-kg object vibrating at the end of a frictionless horizon

Question

The position of an 14.6-kg object vibrating at the end of a frictionless horizontal spring with k = 100 N/m is given by:

X = (0.25 m) cos(5pt/6) [note: cosine(5 x 3.14 x t / 6)]

1. Find the period [sec].
0.5 < T < 2.5
2.5 < T < 5.0
5.0 < T < 1.0
0 < T < 0.5


2. Find the velocity [m/s] of the object at t = 1.0 sec.
0 < v
-3.0 < v < -0.3
-0.03 < v < 0
-0.3 < v < -0.03


3. Find the total energy [J] of the object at t = 1.0 sec.
1.5 < E < 3.0
0 < E < 1.5
3.0 < E < 4.5
4.5 < E < 6.0


4. While standing at a crosswalk, you hear a frequency of 560 Hz from an approaching police car. After the police car passes, its frequency is 480 Hz. What is the speed of the police car? (speed of sound = 340 m/s)
21.1 m/s
26.2 m/s
17.4 m/s
13.1 m/s

Explanation / Answer

Mass m =14.6-kg
Spring constant k = 100 N/m
Position function X = (0.25 m) cos(5t/6)
Compare this with X = A cos (t) we get
Amplitude A = 0.25 m
Angular frequency = 5
(1).The period T =2 /
                          = 0.4
i.e., 0 < T < 0.5
(2). The velocity v = dX / dt
                             = 0.25 x (5 /6)[ -sin (5t/6)]
The velocity [m/s] of the object at t = 1.0 sec is v = 0.25 x (5/6) [ -sin (5(1)/6)]
                                                                            = 0.25 x (5/6) x (-0.5)
                                                                            = -0.327 m / s


i.e., -3.0 < v < 0.3
3). Maximum velocity V = 0.25 x (5/6)
                                       = 0.6544 m / s
The total energy [J] of the object at t = 1.0 sec is E = ( 1/ 2) mV 2
                                                                                                       = 3.127 J
i.e.,3.0 < E < 4.5
4). Speed of sound V = 340 m/ s
Speed of car v = ?
Apperent frequency f ' = 560 Hz
We know f ' = [ V / ( V - v ) ] f
           560 = [ 340 / ( 340 - v ) ] f         ------( 1)
After passes the car :
---------------------
Apperent frequency f " = [ V / ( V+ v) ] f
                           480 = [ 340 / ( 340 +v) ] f            ------( 2)
eq( 1) / eq( 2) ==> (560/480) = [(340+ v) / ( 340 - v) ]
                              1.1666 = [(340+ v) / ( 340 - v) ]
                             340 + v = 1.1666 ( 340 - v)
                                          = 396.66 - 1.1666 v
                          2.1666 v = 56.666
                                      v = 26.15 m / s
                                        ~ 26.2 m/ s

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