# Physics: A Mass is Attached to a Spring and set into Oscillation

**Problem: **A 0.54 kg mass is attached to a light spring with a force constant of 32.9 N/m and set into oscillation on a horizontal frictionless surface. If the spring is stretched 5.0 cm and released from rest, determine the following.

M = 0.54 kg

K = 32.9 N/m

X = 5.0 cm = 0.05 m

**(a)** maximum speed of the oscillating mass

**Solution: **

We know that the conservation of mass equation is:

Etot = KE + PE = MVmax²/2 *PE is 0 when KE is maximized, thus the right side of the equation*

Setting KE = 0 and, thus, maximizing PE, set PE = MVmax²/2

MVmax²/2 = KX²/2

Solve for Vmax to get:

**Answer: Vmax = 0.39 m/s**

**(b)** speed of the oscillating mass when the spring is compressed 1.5 cm

**Solution: **

X = 1.5 cm = 0.015 m

Use the conservation of mass formula from above, plugging in the respective values that we have found. Solve for V (not Vmax) to get:

V = sqrt((m(Vmax²) – KX²)/M)

**Answer: V = 0.372 m/s**

**©** speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position

**Answer: Same as (b)**

**(d)** value of *x* at which the speed of the oscillating mass is equal to one-half the maximum value

**Solution: **

Use the conservation of energy equation from above, substituting V for Vmax/2 and then solving for X

X = sqrt((M(Vmax²) – M(Vmax/2)²)/K)

**Answer: X = 0.043 m**