Question

A toy of mass 0.155 kg is undergoing simple harmonic motion (SHM) on the end of a horizontal spring with force constant 305 N/m. When the object is a distance 1.15 times 10^-2 m from its equilibrium position, it is observed to have a speed of 0.305 m/s. What is the total energy of the object at any point of its motion? What is the amplitude of the motion?

Answer 1

part A

the total mechanical energy is equal to

$E=U_{k}+K=\frac{1}{2}k\cdot x^{2}+\frac{1}{2}\cdot m\cdot v^{2}$   elastic potential energy and kinetic energy

Where:

elastic constant

$k=305\frac{N}{m}$
elongation

$x=1.15\cdot 10^{-2}m$
speed

$v=0.305\frac{m}{s}$
mass

$m=0.155kg$

evaluate numerically

$E=\frac{1}{2}\cdot 305\frac{N}{m}\cdot \left ( 1.15\cdot 10^{-2}m \right )^{2}+\frac{1}{2}\cdot 0.155kg\cdot \left ( 0.305\frac{m}{s} \right )^{2}=0.0274J$

$E=0.0274J$

Part B

The total mechanical energy is equal to

$E=\frac{1}{2}\cdot k\cdot A^{2}$

where:

is the amplitude

expression for the amplitude.

$E=\frac{1}{2}\cdot k\cdot A^{2}\Rightarrow A=\sqrt{\frac{2\cdot E}{k}}$

evaluating numerically

$A=\sqrt{\frac{2\cdot 0.0274J}{305\frac{N}{m}}}=1.34\cdot 10^{-2}m$

part C

mechanical energy is also equal to

$E=\frac{1}{2}\cdot m\cdot v_{max}^{2}$

Expression for the maximum speed

$E=\frac{1}{2}\cdot m\cdot v_{max}^{2}\Rightarrow v_{max}=\sqrt{\frac{2\cdot E}{m}}$

evaluating numerically

$v_{max}=\sqrt{\frac{2\cdot E}{m}}=\sqrt{\frac{2\cdot 0.0274J}{0.155kg}}=0.595\frac{m}{s}$

finally

$v_{max}=0.595\frac{m}{s}$