College

A 0.26 kg mass is attached to a light spring with a force constant of 35.9 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

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

Answer :

Final answer:

The maximum speed of the mass in oscillation is found by converting the potential energy at maximum stretch into kinetic energy, and the speed at a particular compression is found by equating the total energy at that point with the initial total energy.

Explanation:

For an object attached to a spring and set into oscillatory motion on a frictionless horizontal surface, the conservation of mechanical energy principle is applicable. The total mechanical energy of the system remains constant throughout the motion since there are no non-conservative forces, like friction, acting on the system.

Maximum Speed Calculation

The maximum speed of the oscillating mass is achieved when all the potential energy stored in the spring is converted into kinetic energy. This occurs at the equilibrium position of the spring where the spring is neither stretched nor compressed.

The potential energy stored in the spring when stretched can be calculated using the formula for elastic potential energy: PEelastic = 1/2 k x2, where k is the spring constant and x is the displacement from the spring's equilibrium position.

The kinetic energy of the mass at its maximum speed can be calculated using the formula: KE = 1/2 m v2, where m is the mass of the object and v is the speed of the object.

By equating the potential and kinetic energies and solving for v, we find:
1/2 k x2 = 1/2 m v2
v = √x(k/m)

Substituting the given values (k = 35.9 N/m, m = 0.26 kg, and x = 5.0 cm or 0.05 m), we can calculate the maximum speed.

Speed at Compression of 1.5 cm

When the spring is compressed at 1.5 cm, part of the energy is stored as potential energy and the rest is kinetic energy. To find the speed of the mass at this position, we use the fact that the total mechanical energy remains constant:

1/2 k x12 = 1/2 k x22 + 1/2 m v2

Where x1 is the initial stretch (5.0 cm), x2 is the current compression (1.5 cm), and v is the velocity at that compression. We solve for v using the given information.

Answer:

(a) maximum speed of the oscillating mass is 0.588 m/s

(b) speed of the oscillating mass when the spring is compressed 1.5 cm is 0.56 m/s

Explanation:

Given;

mass of the object, m = 0.26 kg

force constant, k = 35.9 N/m

spring displacement, A = 5.0 cm

Part (a) maximum speed of the oscillating mass

Vmax. = ωA

Where;

ω is angular speed

A is the maximum displacement

[tex]\omega= \sqrt{\frac{k}{m}} = \sqrt{\frac{35.9}{0.26}} = 11.75 \ rad/s[/tex]

Vmax. = 11.75 x 0.05 = 0.588 m/s

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

x = 1.5 cm

[tex]V = \sqrt{\frac{k}{m}(A^2-x^2) }\\\\ V = \sqrt{\frac{35.9}{0.26}(0.05^2-0.015^2) } = 0.56 \ m/s[/tex]

V = 0.56 m/s

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