Math Problem Statement

  1. At t=1.00[s], a 0.350 [kg] mass in a horizontal spring mass system with spring constant 175 [N/m] was measured to be 3.00 [cm] away from equilibrium with a speed of 4.50 [cm/s]. The spring was stretched at that time but is getting compressed. Describe the position of the mass as a function of time.

  2. Initially, a 0.600 [kg] mass was used to compress a spring 2.50 [cm] in a horizontal spring mass system by holding the mass, then letting it go. Earlier calibration of the spring earlier showed that the spring extends 3.00 [cm] for a 750 [N] weight. Describe the position of the mass as a function of time.

Solution

Problem 5: Position of the mass as a function of time

Given:

  • Mass m=0.350kgm = 0.350 \, \text{kg}
  • Spring constant k=175N/mk = 175 \, \text{N/m}
  • Displacement at t=1.00st = 1.00 \, \text{s}: x(1.00)=0.030mx(1.00) = 0.030 \, \text{m}
  • Speed at t=1.00st = 1.00 \, \text{s}: v(1.00)=0.045m/sv(1.00) = 0.045 \, \text{m/s}

Step 1: Find angular frequency ω\omega:

ω=km=1750.350=22.36rad/s\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{175}{0.350}} = 22.36 \, \text{rad/s}

Step 2: Form of the solution:

The position as a function of time is: x(t)=Acos(ωt+ϕ)x(t) = A \cos(\omega t + \phi)

where:

  • AA is the amplitude,
  • ω\omega is the angular frequency,
  • ϕ\phi is the phase constant.

Step 3: Use initial conditions to find AA and ϕ\phi:

At t=1.00st = 1.00 \, \text{s}, x(1.00)=0.030=Acos(22.361+ϕ)x(1.00) = 0.030 = A \cos(22.36 \cdot 1 + \phi) and the velocity v(t)v(t) is: v(t)=Aωsin(ωt+ϕ)v(t) = -A \omega \sin(\omega t + \phi) At t=1.00st = 1.00 \, \text{s}, v(1.00)=0.045=A(22.36)sin(22.361+ϕ)v(1.00) = 0.045 = -A (22.36) \sin(22.36 \cdot 1 + \phi)

These two equations can be solved simultaneously to find AA and ϕ\phi.

From these equations:

  • A0.0308mA \approx 0.0308 \, \text{m},
  • ϕ2.18rad\phi \approx -2.18 \, \text{rad}.

Final position equation: x(t)=0.0308cos(22.36t2.18)x(t) = 0.0308 \cos(22.36 t - 2.18)


Problem 6: Position of the mass as a function of time

Given:

  • Mass m=0.600kgm = 0.600 \, \text{kg}
  • Initial compression x0=0.025mx_0 = 0.025 \, \text{m}
  • Force constant kk derived from calibration: the spring extends 3.00 cm for a weight of 750N750 \, \text{N}.

Step 1: Find spring constant kk:

k=ForceExtension=7500.03=25000N/mk = \frac{\text{Force}}{\text{Extension}} = \frac{750}{0.03} = 25000 \, \text{N/m}

Step 2: Find angular frequency ω\omega:

ω=km=250000.600=204.12rad/s\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{25000}{0.600}} = 204.12 \, \text{rad/s}

Step 3: Form of the solution:

The motion is simple harmonic: x(t)=Acos(ωt+ϕ)x(t) = A \cos(\omega t + \phi)

At t=0t = 0, the mass is at its maximum compression: x(0)=x0=A=0.025m,v(0)=0    ϕ=0x(0) = x_0 = A = 0.025 \, \text{m}, \quad v(0) = 0 \implies \phi = 0

Final position equation: x(t)=0.025cos(204.12t)x(t) = 0.025 \cos(204.12 t)


Relative Questions:

  1. What is the relationship between angular frequency, mass, and spring constant in simple harmonic motion?
  2. How do initial conditions (displacement and velocity) determine amplitude and phase?
  3. Why is the phase constant ϕ=0\phi = 0 for problem 6 but not for problem 5?
  4. How does the spring calibration process help in determining the spring constant?
  5. How would damping affect the position equations in these problems?

Tip: Always double-check units for consistency when solving physics problems, especially when calculating spring constants and frequencies.

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Math Problem Analysis

Mathematical Concepts

Simple Harmonic Motion
Trigonometric Functions
Physics of Oscillations

Formulas

Angular frequency: ω = sqrt(k/m)
Position in SHM: x(t) = A cos(ωt + φ)
Velocity in SHM: v(t) = -Aω sin(ωt + φ)

Theorems

Hooke's Law
Equations of Simple Harmonic Motion

Suitable Grade Level

Undergraduate (Physics or Advanced High School Physics)