A toy of mass 0.151 kg is undergoing SHM on the end of a horizontal spring with force constant k = 301 N/m . When the object is a distance 1.23×10^−2 m from its equilibrium position, it is observed to have a speed of 0.297 m/s.
A-What is the total energy of the object at any point of its motion?
B-What is the amplitude of the motion?
C-What is the maximum speed attained by the object during its motion?
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To find the answers to the given questions, we need to use the concept of simple harmonic motion (SHM) and the equations related to it.
A- To find the total energy of the object at any position in SHM, we need to use the equation:
E = (1/2)kx^2 + (1/2)mv^2
Where:
E is the total energy
k is the force constant of the spring
x is the displacement from the equilibrium position
m is the mass of the object
v is the velocity of the object
Given:
k = 301 N/m
m = 0.151 kg
x = 1.23×10^−2 m
v = 0.297 m/s
Plugging in these values into the equation, we can calculate the total energy:
E = (1/2)(301 N/m)(1.23×10^−2 m)^2 + (1/2)(0.151 kg)(0.297 m/s)^2
B- The amplitude of the motion can be found using the equation:
A = x + (1/2)x^2 / (2k/m)
Where:
A is the amplitude of the motion
Plugging in the given values:
A = (1.23×10^−2 m) + (1/2)(1.23×10^−2 m)^2 / (2)(301 N/m) / (0.151 kg)
C- The maximum speed attained by the object during its motion occurs at the amplitude position. So, the maximum speed can be found using the equation:
v_max = ωA
Where:
v_max is the maximum speed
ω is the angular frequency of the motion
A is the amplitude of the motion
The angular frequency can be calculated using the equation:
ω = √(k/m)
Plugging in the given values, we can find the maximum speed attained by the object during its motion.