Question

A spring has a force constant k and mass m. The spring hangs vertically and a block of unknown mass is attached to its bottom end. It is known that the mass of the block is much greater than that of the spring . The hanging block stretches the spring twice its relaxed length. How long (t) would it take for a low amplitude transverse pulse to travel the length of the stretched spring –

• A. $\sqrt{\frac{2m}{k}}$
• B. $\sqrt{\frac{m}{2k}}$
• C. $\sqrt{\frac{m}{k}}$
• D. None

Answer 1

Answer: (A)

$\sqrt{\frac{2m}{k}}$

Answer 2

Since mass of spring is small compared to the mass M, the tension force is approximately constant

T = Mg

kx = Mg ̃ x = $\frac{Mg}{k}$

where x = elongation of the spring

x = L given.

where L = Relaxed length

New length= 2L = $\frac{2Mg}{k}$

\ mass per unit length

µ = $\frac{mk}{2Mg}$

speed of wave V = $\sqrt{\frac{T}{µ}}$

= $\sqrt{\frac{Mg \times 2Mg}{mk}}$

= Mg $\sqrt{\frac{2}{mk}}$ = Vx $\sqrt{\frac{2}{mk}}$= L $\sqrt{\frac{2k}{m}}$

time = $\frac{2L}{V}$= $\frac{2L\sqrt{m}}{L\sqrt{2k}}$ = $\sqrt{\frac{2m}{k}}$