The mass per unit length of a uniform wire is 0.135 g/cm. A transverse wave of the form y = - 0.21 sin (x + 30t) is produced in it, where x is in meter and t is in second. Then, the expected value of tension in the wire is x × 10⁻² N. Value of x is _________. (Round off to the nearest integer)

Solution

Physics sir · 4 steps

Read the wave speed straight off the equation as ω/k, then tension is just μ v².

1. Wave speed. Compare y = -0.21 sin(x + 30t) with y = A sin(kx + ω t). So k = 1 m⁻¹ and ω = 30 rad s⁻¹, which gives v = ω/k = 30 m s⁻¹.
2. Convert to SI. μ = 0.135 g cm⁻¹ = (0.135 × 10⁻³ kg)/(10⁻² m) = 1.35 × 10⁻² kg m⁻¹.

   v = √(T/μ) ⇒ T = μ v²

3. Put in numbers. T = 1.35 × 10⁻² × 30² = 1.35 × 10⁻² × 900 = 12.15 N.
4. So. T = 12.15 N = 1215 × 10⁻² N, so x = 1215.

The common slip: 12150. Using μ = 0.135 directly as if it were in kg m⁻¹ gives T = 121.5 N, that is 12150, not 1215. One g/cm is 0.1 kg m⁻¹.

The idea to keep. For any wave written as sin(kx ± ω t), the speed is ω/k. On a string, tension is μ v², with μ always in kg per metre first.

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