Problem Statement
Solve the oscillation/wave problem: Solve the oscillation/wave problem: Wire: wave speed 200 m/s at tension 200 N. Tension increased to 800 N. New speed? $v\propto\sqrt{T}$ Step 1: $v\propto\sqrt{T}$; $v_2=200\sqrt{800/200}=200\times2=400$ m/s. $$\boxed{v_2=400\text{ m/s}}$$ Mass $m$ and spring constant $k$ (or equivalent), or wave pa
Given Information
- $\boxed{v_2=400\text{ m/s}$
Physical Concepts & Formulas
Newton’s second law $\mathbf{F}_\text{net} = m\mathbf{a}$ is the fundamental relation between net force and acceleration. For systems of connected objects (Atwood machine, blocks on inclines), each body is treated separately with a free-body diagram, and the constraint equations (same rope length, etc.) link the accelerations.
- $\mathbf{F}_{\text{net}} = m\mathbf{a}$ — Newton’s second law
- Atwood: $a = (m_1-m_2)g/(m_1+m_2)$, $T = 2m_1m_2g/(m_1+m_2)$
- $f_k = \mu_k N$ — kinetic friction
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\boxed{v_2=400\text{ m/s}}$$
$$T = 2\pi\sqrt{\frac{m}{k}}\quad,\quad v_{\max} = A\omega_0 = A\sqrt{\frac{k}{m}}$$
$$\boxed{T = 2\pi\sqrt{m/k}}$$
Answer
$$\boxed{v_2=400\text{ m/s}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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