A uniform rope of length 10 m and mass 15 kg hangs vertically from a rigid support. A block of mass 5 kg is attached to the free end of the rope. If a transverse pulse of wavelength 0.04 m is produced at the lower end of the rope, the wavelength of the pulse when it reaches the top of the rope will be

CONCEPT:

• Simple Harmonic Motion (SHM): Simple harmonic motion is a special type of periodic motion or oscillation where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.
  • Example: Motion of an undamped pendulum, undamped spring-mass system.
• The speed of a transverse wave on a stretched string is given by:

\({\rm{v}} = \sqrt {\frac{{\rm{T}}}{{\rm{\mu }}}}\)

Where v = the velocity of the wave, T = the tension in the string, and μ = mass per unit length.

• The wavelength of the wave on the string is given by,

\(λ = \frac{v}{f}\)

EXPLANATION:

Given - Tension at the lower end (T₁) = 5 kg, Tension at the top end (T₂) = 5 kg + 15 kg = 20 kg and wavelength (λ₁) = 0.04 m

• The speed of transverse waves on the lower end of the stretched string is

\(⇒ {\rm{v_1}} = \sqrt {\frac{{\rm{T_1}}}{{\rm{\mu }}}}\)    ------- (1)

• The speed of transverse waves on the top end of the stretched string is

\(⇒ {\rm{v_2}} = \sqrt {\frac{{\rm{T_2}}}{{\rm{\mu }}}}\)    ------- (2)

On dividing equation 1 and 2, we get

\(⇒ \frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{5}{20}}=\frac{1}{2}\)

⇒ v₂ = 2v₁

As velocity (v) = frequency (f) × wavelength (λ)

⇒ fλ2 = 2fλ1

As frequency remains the same, therefore 

⇒ λ2 = 2λ1

⇒ λ2 = 2 × 0.04 = 0.08 m