PYQ NEET- Concept Of Waves And Electromagnetic Waves

• (2014)

A string of length 1.5 m is fixed at both ends. It is vibrating in its third harmonic and the amplitude at the antinodes is 4 cm. The amplitude of the particle at a distance of 27 cm from one of the ends will be:

The answer is 2 cm.

The amplitude of the particle at a distance of x from one of the ends is given by:

A = A0 * sin(2πx/λ)

where,

• A0 is the amplitude of the antinodes
• x is the distance from one of the ends
• λ is the wavelength

In this case, the wavelength is λ = 2L/3 = 2 * 1.5 m = 3 m.

The amplitude of the particle at a distance of 27 cm from one of the ends is:

A = A0 * sin(2π * 27 cm/3 m) = A0 * sin(π/2) = A0 * 1 = 4 cm * 1 = 4 cm