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Mass-Spring System: Natural Frequency, Period, and Maximum Speed

Created by @nathanedwards

at November 2nd 2023, 4:02:11 pm.

AP_Physics_1-Questions

#displacement

#oscillation

#mass-spring-system

Question:

A mass-spring system consists of a spring with spring constant $k = 200 \ \text{N/m}$ and a mass $m = 2 \ \text{kg}$ . The spring is stretched by 0.4 m from its equilibrium position and then released. The system is free from any external forces or damping. Assume the positive direction is upward.

Determine the natural frequency of oscillation for this system.
How much time does it take for the mass to complete one full oscillation?
Determine the maximum speed of the mass during its motion.

Answer:

Given: Spring constant, $k = 200 \ \text{N/m}$ Mass, $m = 2 \ \text{kg}$ Displacement from equilibrium position, $x = 0.4 \ \text{m}$ Positive direction is upward

1. Determine the natural frequency of oscillation for this system.

The natural frequency of a mass-spring system is given by the equation:

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

Substituting the given values,

f = \frac{1}{2\pi}\sqrt{\frac{200}{2}}

f = \frac{1}{2\pi}\sqrt{100}

f = \frac{1}{2\pi} \cdot 10

f = \frac{5}{\pi} \ \text{Hz}

Thus, the natural frequency of oscillation for this system is $\frac{5}{\pi}$ Hz.

2. How much time does it take for the mass to complete one full oscillation?

The period of oscillation for a mass-spring system is given by the equation:

T = \frac{1}{f}

Substituting the natural frequency calculated in the previous part,

T = \frac{1}{\frac{5}{\pi}}

T = \frac{\pi}{5} \ \text{s}

Thus, it takes $\frac{\pi}{5}$ seconds for the mass to complete one full oscillation.

3. Determine the maximum speed of the mass during its motion.

The maximum speed of a mass undergoing simple harmonic motion is given by the equation:

v_{\text{max}} = A\omega

where $A$ is the amplitude and $\omega$ is the angular frequency.

Given that the displacement from the equilibrium position, $x = 0.4 \ \text{m}$ , which is equal to the amplitude $A$ . We already calculated the natural frequency $f$ in the first part, which can be converted to angular frequency $\omega$ using the relation $\omega = 2\pi f$ .

\omega = 2\pi \left(\frac{5}{\pi}\right)

\omega = 10\pi \ \text{rad/s}

Substituting these values,

v_{\text{max}} = (0.4)(10\pi)

v_{\text{max}} = 4\pi \ \text{m/s}

Therefore, the maximum speed of the mass during its motion is $4\pi$ m/s.