3. A 0.66 kg mass is attached to a light spring with a force constant of 24.9 N/
ID: 1449197 • Letter: 3
Question
3. A 0.66 kg mass is attached to a light spring with a force constant of 24.9 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.
(a) maximum speed of the oscillating mass
_____________m/s
(b) speed of the oscillating mass when the spring is compressed 1.5 cm
____________m/s
(c) speed of the oscillating mass as it passes the point 1.5 cm from the equilibrium position
___________m/s
(d) value of x at which the speed of the oscillating mass is equal to one-half the maximum value
___________m
Explanation / Answer
M = 0.66 kg
K = 24.9 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 = M(Vmax)2 / 2 PE is 0 when KE is maximized, thus the right side of the equation
Setting KE = 0 and, thus, maximizing PE, set PE = M(Vmax)2 / 2
M(Vmax)2 / 2 = KX² / 2
Solve for Vmax to get:
Vmax = sqrt(KX² / M)
Vmax = sqrt(24.9*0.052 / 0.66)
Answer: Vmax = 0.31 m/s
(b) speed of the oscillating mass when the spring is compressed 1.5 cm
Solution:
X = 1.5 cm = 0.015 m
total E = Umax = ½kX² = ½ * 24.9N/m * (0.05m)² = 0.0311 J
At x = 0.015 m, U = ½ * 24.9N/m * (0.015m)² = 0.0028 J
leaving KE =0.0311J -0.0028J = 0.0283 J = ½mv² = ½ * 0.66kg * v²
v = 0.29 m/s
Answer: V = 0.29 m/s
(c) 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)
X = sqrt[(0.66*0.312) - 0.66*0.1552] / 24.9)
Answer: X = 0.0437 m
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