Problem Statement
Solve the oscillation/wave problem: Solve the oscillation/wave problem: String with $\mu_0$ over frictionless pulley, mass $m$ hanging. Find wave speed on horizontal portion. $T=mg$ from hanging mass; $v=\sqrt{T/\mu}$ Step 1: Tension $T=mg$; $v=\sqrt{mg/\mu_0}$. $$\boxed{v=\sqrt{mg/\mu_0}}$$ Mass $m$ and spring constant $k$ (or equiva
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Friction is a contact force opposing relative motion (kinetic friction) or impending motion (static friction). On an inclined plane, the weight component along the slope is $mg\sin\theta$ and the normal force is $N = mg\cos\theta$, giving maximum static friction $f_{s,\max} = \mu_s mg\cos\theta$. The condition for sliding is $\tan\theta > \mu_s$.
- $f = \mu N$ — kinetic friction force
- $N = mg\cos\theta$ — normal force on incline
- $mg\sin\theta – \mu mg\cos\theta = ma$ — Newton’s 2nd law along incline
- $\tan\theta_c = \mu_s$ — critical angle for sliding
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
$$\boxed{v=\sqrt{mg/\mu_0}}$$
$$T = 2\pi\sqrt{\frac{m}{k}}\quad,\quad v_{\max} = A\omega_0 = A\sqrt{\frac{k}{m}}$$
$$\boxed{T = 2\pi\sqrt{m/k}}$$
Answer
$$\boxed{v=\sqrt{mg/\mu_0}}$$
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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