Physics · Class 11

Identify string nodes and harmonics

Two opposite traveling waves form a standing pattern with fixed nodes. Both fixed ends must be nodes, so only wavelengths 2L/n fit. Higher tension raises all harmonic frequencies; greater linear density lowers them.

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Identify string nodes and harmonics

இரு நிலையான முனைகளும் கணுக்கள். L நீள கம்பியில் λ = 2L/n என்ற அலைநீளங்களே பொருந்தும்.

Key notes, relationships and model limits

Key relationship

v=√(T/μ); fn=nv/(2L); y=2A sin(nπx/L)cos(2πfnt).

Remember

Common trap

A standing wave has fixed nodes rather than a shape translating along the string.

Model limits

Uniform string fixed at both ends; small-amplitude standing wave with imposed total amplitude 0.15 m for visualization. High frequencies require time scrubbing to inspect a phase and may alias during live animation.

What you will learn

Connect boundary nodes, harmonic number and tension-dependent wave speed.

Try the experiment

Try both scenarios, change one control, and predict the result before checking the graph and readouts.

Think of it this way

Change one input while keeping the others fixed, as you would in a controlled experiment.

Worked example

For T=40 N, μ=0.01 kg/m and L=1 m, v=63.246 m/s. The second harmonic has f=63.246 Hz and λ=1 m.

Why this matters

The model connects identify string nodes and harmonics with an observable reading. Check the equation and assumptions before applying it.

Before you start

SI units · algebra · graph reading

Check your understanding

For the second harmonic of a string fixed at both ends, how many interior nodes occur?

Two interior nodes
One interior node
No interior nodes

Read the explanation

n=2 gives nodes at x=0, L/2 and L, with one inside the string.

Course connection: Class XI Chapter 14. Consult the roadmap for CBSE, NEET, JEE Main and Advanced scope.