Given :
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface.
The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.
To Find :
At what location are the kinetic energy and the potential energy the same.
Solution :
Let, at location x from the equilibrium position the kinetic energy and the potential energy the same.
So,
[tex]P.E = K.E\\\\\dfrac{kx^2}{2} = \dfrac{mv^2}{2}\\\\x = v\sqrt{\dfrac{m}{k}}\\\\x = 1.5 \times \sqrt{\dfrac{0.5}{20}}\ m\\\\x = 0.238 \ m[/tex]
Hence, this is the required solution.
