Physics · Waves · NEET
A horizontal light string has the same tension everywhere. But a vertical rope has weight. Any point in the rope must support everything below it. Near the bottom there is little rope below, so tension is small. Near the top the whole rope (and any hanging block) pulls down, so tension is large. That is why tension increases with height x above the lowest point.
It speeds UP. Wave speed is v = sqrt(T/mu). As the pulse climbs, tension T increases (more rope hangs below it), and the linear mass density mu stays the same for a uniform rope. So v increases with height. This is a very common NEET trap: students assume constant speed like on a normal string.
Linear mass density is mu = m/L. The mass of rope below height x is mu*x = (m/L)*x. That segment's weight is the tension holding it: T(x) = (m/L)*x*g = mu*g*x. So at the bottom (x = 0) tension is 0 (ideal case), and at the top (x = L) tension is m*g, the full rope weight.
The frequency of the pulse stays the same everywhere (the source sets it, and it does not change along the rope). Since v = f*lambda and v increases upward, lambda must also increase: lambda is proportional to v, which is proportional to sqrt(T). So lambda2/lambda1 = sqrt(T_top/T_bottom). This is exactly the NEET 2016 question.
Then the bottom is not tension-free. At the very bottom, tension = m2*g (it holds the block). At the top, tension = (m1 + m2)*g (it holds the whole rope of mass m1 plus the block). Use these two tensions in v = sqrt(T/mu) and the ratio of speeds (or wavelengths, since f is constant) is sqrt(T_top/T_bottom).
A uniform rope of length L and mass m1 hangs vertically from a rigid support, with a block of mass m2 attached to its free (lower) end. A transverse pulse of wavelength lambda1 is produced at the lower end of the rope; when it reaches the top its wavelength is lambda2. The ratio lambda2/lambda1 is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
v = sqrt(T/mu), where T is the tension at that point and mu is the linear mass density. For a uniform rope with a free bottom end, tension at height x is T = mu*g*x, so v = sqrt(g*x). The speed depends on position along the rope.
For an ideal uniform rope with nothing attached at the free lower end, tension there is zero, so v = sqrt(T/mu) gives zero speed right at the tip. In practice a small pulse still moves, but the idealized formula gives v = 0 at x = 0 and grows as sqrt(x) upward.
No. For a uniform rope, mu = m/L is the same at every point. Only the tension changes with height. That is what makes the wave speed vary, not mu.
The source sets the frequency, and frequency is not altered as a wave travels through one medium. Since v = f*lambda and v increases up the rope while f is fixed, the wavelength lambda must increase to keep the product equal to the new v.
