Question

a uniform rope of length L and mass m1 hangs vertically from a rigid support a block of mass m2 is attached to the free end of the Rope the transverse pulse of wavelength λ1 is produced at the lower end of the Rope the wavelength of the pulse when it reaches the top of the Rope is λ2 the ratio λ2/λ1 is​

Answer 1

Answer:

Wavelength of pulse at the lower end,

"refer the attachment "

Answer 2

S O L U T I O N

According to the question, we have

Wavelength of transphase pulse

λ = v/f _[i]

( v = velocity of the wave ; f = frequency of the wave )

As we know : v = √T/μ _[ii]

( T= tension in the spring ; μ= mass per unit length of the rope )

From Eq [i] and [ii] we get

λ = 1/f √T/μ => λ ∝ √T

So, for two different case, we get

λ2/λ1 = √T2/T1 = √[(m1 + m2) / m2]

∴ The ratio λ2/λ1 is √m1+m2 / m2 !