An object with mass 3.3 kg is executing simple harmonic motion, attached to a spring with spring constant 260 N/m . When the object is 0.017 m from its equilibrium position, it is moving with a speed of 0.40 m/s . Part A Calculate the amplitude of the motion. Express your answer to two significant figures and include the appropriate units.

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 = 1/2kx²1

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

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