Answer: the amplitude of the motion is 0.048 m
Explanation:
Given that;
mass m = 3.3 kg
spring constant k = 260 N/m
When the object is x = 0.017 m from its equilibrium position,
it is moving with a speed of v = 0.40 m/s
amplitude of the motion = ?
we know that;
formula for Potential of the string PE = 1/2kx²------1
Kinetic energy KE = 1/2mv² --------2
Now expression for the maximum potential energy of the spring is
PE_max = 1/2 kx²_max -----3
Now from conservation of Energy
PE_spring = PE + KE
from our equation 1, 2 and 3
1/2kx²_max = 1/2kx² + 1/2mv²
now we rearrange for x_max
x²_max = (1/2kx² + 1/2mv²) / 1/2k
x²_max = (kx² + mv²) / k
x_max = √ [(kx² + mv²) / k]
now we substitute our values into the equation
x_max = √ [(260(0.017)² + 3.3×(0.40)²) / 260]
= √((0.07514 + 0.528) / 260)
= √(0.60314 / 260)
= √ 0.002319
x_max = 0.048 m
Therefore the amplitude of the motion is 0.048 m
