Question

A mass of 2 kg is suspended by a string of mass 6 kg. A wave of wavelength 6cm is produced at the bottom of string. the wavelength of wave at the top end of string will be​

Answer 1

Answer:

Let us assume the length of the string is 'l' m

Now calculating the mass density of the string we get:

$\mu = \dfrac{Mass}{Length}\\\\\boxed{\mu = \dfrac{8}{l}}$

Now let us draw the free body diagram of the given setup.

(Refer to the attachment)

Now, since the string has mass, the tension would be different near the mass of 2 kg (lower end of the string)

Tension at the lower end is:

T = 2g ( g = 9.8 m/s² )

We know that, velocity of a wave in a string is given by:

$\boxed{v = \sqrt{ \dfrac{T}{\mu}}}$

Also it is known that velocity of wave is:

→ v = λ × f

Hence equating these two we get:

$\rightarrow \sqrt{ \dfrac{T}{\mu}} = \lambda \times f\\\\\\\text{According to question, wavelength is 6 cm which is 0.06 m}\\ \text{Substituting the values we get:}\\\\\rightarrow \sqrt{ \dfrac{2g}{8/l}} = 0.06 \times f\\\\\rightarrow \sqrt{\dfrac{2gl}{8}} = 0.06 \times f\\\\\rightarrow f = \dfrac{1}{0.06}\times\sqrt{gl}\times\dfrac{1}{2}\\\\\\\rightarrow f = \dfrac{ \sqrt{gl}}{0.12} \:\: \rightarrow (Eqn.\:1)$

Now we know that Tension at the top of the string is 8g.

Hence calculating the velocity we get:

$\rightarrow v = \sqrt{\dfrac{ 8g}{8/l}}\\\\\rightarrow v = \sqrt{gl}$

Hence velocity at the top end would be √gl.  Since the frequency is constant, we get:

→ λ = v / f

$\rightarrow \lambda = \dfrac{\sqrt{gl}}{\dfrac{\sqrt{gl}}{0.12}}}\\\\\\\rightarrow \lambda = \sqrt{gl} \times \dfrac{0.12}{\sqrt{gl}}\\\\\boxed{ \bf{\lambda = 0.12 m}}$

Hence the wavelength at the top end would be 12 cm.