Question

A mass of 3.6 kg oscillates on a horizontal spring with a spring constant of 160 N/m. When the mass passes the equilibrium point its speed is 5.2 m/s. d. What is the maximum displacement of the mass?

Answer 1

Answer:

x = 0.78 m = 78 cm

Explanation:

Applying the law of conservation of energy to the spring-mass system, we can write the following equation:

$Kinetic\ Energy\ at\ Mean\ Position= Potential\ Energy\ at\ extreme\ position\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2\\\\x^2 = \frac{mv^2}{k}\\\\x = \sqrt{\frac{mv^2}{k} }$

where,

x = maximum displacement = ?

m = mass = 3.6 kg

v = speed at mean position = 5.2 m/s

k = spring constant = 160 N/m

Therefore,

$x = \sqrt{\frac{(3.6\ kg)(5.2\ m/s)^2}{160\ N/m} }$

x = 0.78 m = 78 cm