JEE Main 2021 · 26 August · morning · Numerical answer · +4 / −0
Two travelling waves produces a standing wave represented by equation,
y = 1.0 mm cos(1.57 cm⁻¹) x sin(78.5 s⁻¹)t.
The node closest to the origin in the region x > 0 will be at x = .............. cm.
Answer and solution
Answer 1
Solution
Physics sir · 3 steps
Here the x-part is a cosine, so the origin is an antinode and the first node sits a quarter wavelength away.
- Read the amplitude. The amplitude at position x is 1.0 mm × cos(kx) with k = 1.57 cm⁻¹. A node is a point where this amplitude is zero at all times.
- Node condition. cos(kx) = 0 when kx = π/2, 3π/2, … The smallest positive one, kx = π/2, gives the node closest to the origin.
x = π/2k = 3.14/(2 × 1.57 cm⁻¹) = 1 cm
- So. Since 1.57 = π/2, the wavelength is λ = 2π/k = 4 cm. The first node is at λ/4 = 1 cm.
The common slip: 2. Using the sine-form node condition kx = π gives x = 2 cm, but for a cosine that point is an antinode, not a node.
The idea to keep. Look at the x-part first. With sin kx, nodes are at kx = nπ. With cos kx, the origin is an antinode and nodes are at kx = π/2, 3π/2, …
Answer from the NTA final answer key. Solution by PrepWiser.