Question
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The position of a particle is given by the expression x = 6.00 cos (2.00pi t + pi/4), where x is in meters and t is in seconds. Determine the frequency. Hz Determine period of the motion. Determine the amplitude of the motion. m Determine the phase constant. Determine the position of the particle at t = 0.310 s A 240 g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.100 s. If the total energy of the system is 2.00 J. Find the force constant of the spring Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N/m Find the amplitude of the motion. A 45.0-g object connected to a spring with a force constant of 45.0 N/m oscillates with an amplitude of 8.00 cm on a frictionless, horizontal surface. Find the total energy of the system. We know the spring constant and the maximum displacement of the object. What is the value of the spring potential energy and the kinetic energy instant? mJ Find the speed of the object when its position is 1.10 cm. (Let 0 cm be the position of equilibrium.) Find the kinetic energy when its position is 3.00 cm. Find the potential energy when its position is 3.00 cm. A 1.70-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 24.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations. Find the force constant of the spring. N/m Find the frequency of the oscillations. Hz Find the maximum speed of the object. m/s Where does this maximum speed occur? x = plusminus Find the maximum acceleration of the object. m/s2 Where does the maximum accleration occur? x = plusminus Find the total energy of the oscillating system. J Find the speed of the object when its position is equal to one-third of the maximum value. m/s Find the magnitude of the acceleration of the object when its position is equal to one-third of the maximum value. m/s2
Explanation / Answer
1) a) 1 Hz
b) 1 sec
c) 6 m
d) Phase constant = (pi)/4 rad.
e) x= -5.5065 m
2) a) Spring constant = 947.48 N/m
b) amplitude = 0.0649 m
3) a) Energy of system = Max. energy of spring = 0.5*k*A*A
= 0.144 J
= 144 mJ = E
b) E = PE + KE
= 0.5*m*v*v + 0.5*k*x*x
From this , we can get "v",
v= 2.505 m/sec
c) KE = 123.75 mJ
d) PE = 0.5*k*x*x = 20.25 mJ
4) a) k*x = F
k= 24/(0.200)
= 120 N/m
b) w = sqrt(k/m)
f = w/2(pi)
= 1.3371 Hz (approx.)
c) Max KE = 0.5*k*A*A
SO, Max. V =sqrt( 48/17) m/sec
= 1.68033 /sec
d) Max.Speed occurs at equi. position , since PE = 0 there,
So, max. speed occurs at x= 0 m
e) Max Force = k*A
So, max acc. occurs at extremes,
Max. Acc = 24/17 m/sec2
= 1.4117 m/s2
f) Max. Acc occurs at x= +- 0.200 m
g) Max. KE = Max. PE = TE
TE = 2.4 J
h) v = 1.584 /sec
i) F= k*x
a= 80/17 m/s2 = 4.7058 m/s2
