A wire of length 30 cm, stretched between rigid supports, has it's nᵗʰ and (n + 1)ᵗʰ harmonics at 400 Hz and 450 Hz, respectively. If tension in the string is 2700 N, it's linear mass density is ____________ kg/m.

Solution

Physics sir · 4 steps

Two neighbouring harmonics differ by exactly one fundamental, so 450 minus 400 gives you it right away.

1. Find the fundamental. For a string fixed at both ends the harmonics are fₙ = n f₁. So two consecutive harmonics differ by f₁: f₁ = 450 - 400 = 50 Hz.

   f₁ = v/2L = 1/2L√(T/μ)

2. Wave speed. With L = 0.3 m, v = 2L f₁ = 0.6 × 50 = 30 m s⁻¹.
3. Solve for μ. From v = √(T/μ), μ = T/v² = 2700/900 = 3 kg m⁻¹.
4. So. The linear mass density of the wire is 3 kg/m.

The common slip: 12. Writing f₁ = v/L instead of v/2L gives v = 15 m s⁻¹ and μ = 12. The fundamental fits half a wavelength in the string, so λ₁ = 2L.

The idea to keep. For a string fixed at both ends, the gap between any two consecutive harmonics is the fundamental frequency. Then use f₁ = v/2L and v = √(T/μ).

Answer from the NTA final answer key. Solution by PrepWiser.