Question

You hang a 200 g mass from a spring and it stretches by 10 cm. You stretch it further by 5 cm and release it to oscillate.

What is the frequency?

Write an equation for the motion.

What is the maximum velocity?

What is the total enrgy in the system (kinetic + spring potential)?

Answer 1

Let 'k' be the spring constant. Then,

k * (0.1 m) = (0.2 kg) * (9.8 m/s²)

=> k = 19.6 N/m

When streched further by 5 cm, it results in a Simple Harmonic Motion (SHM). Equation of motion is given by:

y = Asin(ωt + φ)

where A is amplitude, ω is angular frequency of motion and φ is initial phase.

Let's consider the coordinate system in which the mean position (y = 0) is the position where spring is streched by 10 cm. Downward movement is negative and upward movement is positive.

So, at t = 0, y = -0.05 m. Hence, φ = -π/2

A = 0.05 m, ω = (k/m)^1/2 = (19.6/0.2)^1/2 = 9.9 rad/s

Frequency of motion, f = ω/2π = 9.9/2π = 1.58 Hz

Hence, Equation of motion is: y = 0.05 sin(9.9t - π/2)

Energy of the system is conserved, so,

Total energy at the mean position = Total energy at the extreme position

=> KE_max = PE_max

=> mv_max²/2 = kA²/2

=> v_max = A(k/m)^1/2

=> maximum velocity, v_max = 0.05 * (19.6/0.2)^1/2 = 0.495 m/s

Total energy of the system is,

E = KE + PE = KE_max = PE_max = kA²/2 = 19.6*0.05²/2 = 0.0245 N