Math Problem Statement

Considering the application of the 2nd law to the hanging mass and cart, what is the acceleration of the cart in terms of the masses of these objects, the spring constant k, and the displacement of the cart from its equilibrium position (i.e., x = 0 when the system is static)? Take care again as the acceleration of the cart would be in the +x direction when the position of the cart (x) is negative. If the mass of the cart m1 = 0.250 kg, the mass of the hanging weight m2 = 0.061 kg, and the spring constant equals 4.1 N/m, what would be the period of oscillation of this system?

Solution

Let’s address each question in turn.


First question:
We are looking for the acceleration aa of the cart in terms of the masses m1m_1 and m2m_2, the spring constant kk, and the displacement xx.

  • The system consists of a spring and a cart. The force exerted by the spring is F=kxF = -kx.
  • Newton's second law gives:
    Fnet=(m1+m2)aF_{\text{net}} = (m_1 + m_2)a, where m1m_1 is the mass of the cart and m2m_2 is the mass of the hanging weight.

Equating the force from the spring and the net force:
kx=(m1+m2)a-kx = (m_1 + m_2)a

Solving for aa:
a=km1+m2xa = -\frac{k}{m_1 + m_2}x

Correct answer:
a=km1+m2x\boxed{a = -\frac{k}{m_1 + m_2}x}


Second question:
We are tasked with calculating the period of oscillation.

The period of oscillation for a spring-mass system is given by:
T=2πmeffkT = 2\pi \sqrt{\frac{m_{\text{eff}}}{k}}

Here, meffm_{\text{eff}} is the effective mass, which for a system with a cart and a hanging mass is:
meff=m1+m2m_{\text{eff}} = m_1 + m_2

Substitute the given values:

  • m1=0.250kgm_1 = 0.250 \, \text{kg}
  • m2=0.061kgm_2 = 0.061 \, \text{kg}
  • k=4.1N/mk = 4.1 \, \text{N/m}

meff=0.250+0.061=0.311kgm_{\text{eff}} = 0.250 + 0.061 = 0.311 \, \text{kg}

T=2π0.3114.1T = 2\pi \sqrt{\frac{0.311}{4.1}}

Simplify:
T=2π0.07585T = 2\pi \sqrt{0.07585}

T=2π(0.2754)T = 2\pi (0.2754)

T1.73secondsT \approx 1.73 \, \text{seconds}

Final answer:
1.73seconds\boxed{1.73 \, \text{seconds}}


Would you like additional details or explanations? Here are five related questions you might find helpful:

  1. How does the effective mass impact the oscillation period?
  2. Why is the displacement xx directly proportional to acceleration in harmonic motion?
  3. Can the spring constant kk change for different materials?
  4. What is the significance of the negative sign in the acceleration equation?
  5. How does adding more mass to the cart affect the oscillation period?

Tip: Always confirm whether a spring obeys Hooke’s law (F=kxF = -kx) for small displacements before applying these equations!

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law
Simple Harmonic Motion
Oscillation Period

Formulas

F = -kx
F_net = (m1 + m2)a
a = -k/(m1 + m2)x
T = 2π√(m_eff/k)

Theorems

Hooke's Law
Newton's Second Law of Motion

Suitable Grade Level

Grades 11-12