- Solution: Velocity of a transverse wave in a
**stretched string**v = μT. . where T is the tension of the**string**and μ is mass per unit length of the**string**. μ = πr2 ×ρ. where r is the**radius**or the**string**and ρ is the density of the**material**of the**string**. ∴ v = r1..**Two strings**of identical**material**and**radius**are**stretched**with the**same**tension with their ends fixed, but one**string**... **Two****strings**A and B, made**of same****material**, are**stretched**by**same**tension. The**radius**of**string**A is double of the**radius**of B.A transverse wave travels on A with speed v A and on B with speed v B. The ratio (v A v B) is. v = √ T μ; as T is constant, v ∝ 1 √ μ. But μ = m a s s l e n g t h = π r 2 ρ → μ ∝ r 2. ⇒ v ∝ 1 r → v ...**Two**wires A and B are of the**same****material**. Their lengths are in the ratio 1 :2 and the diameter are in the ratio 2:1. If they are pulled by the >**same**force, then increase in length will be in the ratio Aptitude Foxoyo Bot now Was your question answered?.- Answer (1 of 5): The speed of a standing wave on a closed-end
**string**is given by: v=\sqrt{\frac{T}{\mu}} where \mu is the linear density of the**string**. Using the frequency, wavelength, speed relation, we get: f=\frac{1}{\lambda}\sqrt{\frac{T}{\mu}} As long as you stay within one harmonic, th. the farm wife ... **Two****strings**A and B, made**of same****material**, are**stretched**by**same**tension. The**radius**of**string**A is double of the**radius**of B. A transverse wave travels on A with speed v A and on B with speed v B. The ratio v A /v B is (a) 1/2 (b) 2 (c) 1/4 (d) 4. 8. Both the**strings**, shown in figure (15-Q1), are made**of same****material**and have**same**