Question

An object with a mass of 0. 25 kg is undergoing simple harmonic motion at the end of a vertical spring with a spring constant, k = 450 N/m. The object is determined to have a velocity of 0.3 m/s when passing through the equilibrium.

1. Find the amplitude of the motion

2. Find the total energy of the object at any point of its motion​

Answer 1

Answer:

1) The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object at any point of its motion is 16.892 joules.

Explanation:

1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy ($U_{e}$), in joules, is equal to maximum translational kinetic energy ($K$), in joules:

$U_{e} = K$

$\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}$ (1)

Where:

$k$ - Spring constant, in newtons per meter.

$A$ - Amplitude, in meters.

$m$ - Object mass, in kilograms.

$v$ - Speed of the object at equilibrium, in meters per second.

If we know that $k = 450\,\frac{N}{m}$, $m = 0.25\,kg$ and $v = 0.3\,\frac{m}{s}$, then the amplitude of the motion is:

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

$k\cdot A^{2} = m\cdot v^{2}$

$A = v\cdot \sqrt{\frac{m}{k} }$

$A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }$

$A \approx 0.274\,m$

The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object ($E$), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: ($k = 450\,\frac{N}{m}$, $A \approx 0.274\,m$)

$E = U_{e}$

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

$E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}$

$E = 16.892\,J$

The total energy of the object at any point of its motion is 16.892 joules.