HC Verma Chapter 15 Problem 9 — tension from fundamental frequency of clamped string

Problem Statement

Solve the Newton’s Laws / mechanics problem: Solve the Newton’s Laws / mechanics problem: String: $\mu=0.5$ g/cm, length 60 cm, fundamental frequency 200 Hz. Find tension. $f_1=v/(2L)$; $T=\mu v^2$ Step 1: $\lambda=2L=1.20$ m; $v=f\lambda=240$ m/s. Step 2: $\mu=0.05$ kg/m; $T=\mu v^2=0.05\times57600=2880$ N. $$\boxed{T=2880\text{ N}}$$ Mass(es

Given Information

  • $\lambda=2L$
  • $\boxed{T=2880\text{ N}$

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

$$\boxed{T=2880\text{ N}}$$

$$\sum F_x = ma_x\quad,\quad \sum F_y = ma_y = 0\quad\text{(if no vertical acceleration)}$$

$$a = \frac{(m_2-m_1)g}{m_1+m_2} = \frac{(5-3)\times9.8}{8} = \frac{19.6}{8} = 2.45\,\text{m/s}^2$$

Answer

$$\boxed{T=2880\text{ N}}$$

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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