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

x = 0.78 m = 78 cm

Step-by-step 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\\(1)/(2)mv^2 = (1)/(2)kx^2\\\\x^2 = (mv^2)/(k)\\\\x = \sqrt{(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{((3.6\ kg)(5.2\ m/s)^2)/(160\ N/m) }\\\\`

x = 0.78 m = 78 cm