HC Verma Chapter 15 Problem 4 — tension for second harmonic resonance

Problem Statement

Solve the oscillation/wave problem: Solve the Newton’s Laws / mechanics problem: A string length 20 cm, linear density 0.40 g/cm is driven at 30 Hz. Find tension for resonance in the second harmonic. $n$th harmonic: $L=n\lambda/2$; $T=\mu v^2$ Step 1: Second harmonic: L=lambda, so lambda=0.20 m. Step 2: v=f*lambda=30 0.20=6.0 m/s. Ste

Given Information

  • $ Step 1: Second harmonic: L=lambda, so lambda=0.20 m$

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

$$T = 2\pi\sqrt{\frac{m}{k}}\quad,\quad v_{\max} = A\omega_0 = A\sqrt{\frac{k}{m}}$$

$$\boxed{T = 2\pi\sqrt{m/k}}$$

Answer

$$\boxed{T = 2\pi\sqrt{m/k}}$$

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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