Practice question
Question
A stationary wave on a string fixed at both ends has a frequency of 90 Hz and a wave speed of 36 m/s.
What is the length of the string?
Explanation
**Superposition of nearly equal frequencies** creates resultant y = 2A cos(Δω·t/2) sin(ω_avg·t), envelope frequency Δf/2, beat frequency Δf. Total beats heard in Δt is f_beat·Δt, explaining counting over seconds. Fundamental: v₁ = (v/2L) . 90 = (36/2L) ⇒ 2L = (36/90) ⇒ 2L = 0.4 ⇒ L = 0.2 m . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 0.2 m, illustrating frequency-length-speed interdependence and quantization by boundaries. This aligns with NCERT Class 11 treatment, emphasizing conservation, symmetry and dimensional consistency useful for CBSE, NEET and CUET.
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.