An elastic string has a force constant k and mass m . The string 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 string. The hanging block stretches the string the twice it's the relaxed length. How long (t) would it take for a low amplitude transverse pulse to travel the length of the string stretched by the hanging block? (Assume that the string follows Hook's law)

Question

An elastic string has a force constant k and mass m . The string 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 string. The hanging block stretches the string the twice it's the relaxed length. How long (t) would it take for a low amplitude transverse pulse to travel the length of the string stretched by the hanging block? (Assume that the string follows Hook's law) (A) √(2 m/k) (B) √(m/k) (C) √(m/2 k) (D) √(2 m/3 k)

Tardigrade Admin · Accepted Answer

Since the mass of the string is small compared to the mass

m

, the tension force is approximately constant

∴ T = m g

k x = m g, x = \frac{m g}{k}

x = 2 L

given
Where

L -

Relaxed length.

N e w l e n g t h = 2 L

= \frac{2 m g}{k}

∴

mass per unit length

µ = \frac{mk}{2 m g} = \frac{k}{2 g}

Speed of wave

V = \frac{T}{μ}

= \frac{m g \times 2 m g}{mk}

V = g \frac{2 m}{k}

t im e = \frac{2 L}{V}

= \frac{2 m g}{k} \times \frac{1}{g} \frac{k}{2 m}

t im e = \frac{2 m}{k}

$\frac{2 m}{k}$

$\frac{m}{k}$

$\frac{m}{2 k}$

$\frac{2 m}{3 k}$

k2m​​

km​​

2km​​

3k2m​​

$\frac{2 m}{k}$

$\frac{m}{k}$

$\frac{m}{2 k}$

$\frac{2 m}{3 k}$