Question

A uniform rope having a mass of 6 kg hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of rope. A transverse pulse of wavelength 0.06 m is produced at the lower end of rope. Its wavelength when it reaches the top end of the rope is given by . Find . Options

Answer 1

$\huge{\boxed{\mathtt{\red{ANSWER}}}}$

Vlower/Vupper = √(Tlower/Tupper)

= √(2*9.8/8*9.8)

= √1/4

= 1/2

VL/VU = 1/2 = lemL/lemU

1/2 = 0.06/lemU

lemU = 0.12m (Ans)

(Mark as brainlist)

Answer 2

Answer:

Step-by-step explanation:

Vlower/Vupper = √(Tlower/Tupper)

= √(2*9.8/8*9.8)

= √1/4

= 1/2

VL/VU = 1/2 = lemL/lemU

1/2 = 0.06/lemU

lemU = 0.12m (Ans)