Question

When a particular hanging mass is suspended from the string, a standing wave with two segments is formed. When the weight is reduced by 2.2 kg, a standing wave with five segments is formed. What is the linear density of the string

Answer 1

Solution :

Mass is varied keeping frequency constant.

Wavelength, λ  $=\frac{2l}{n}$

where length of spring = l

           number of segments = n

Velocity, v = λ x f

                 = $\sqrt{\frac{T}{\mu}}$

$\mu$ =  mass density, T = tension in string

$T=\frac{4 \mu l^2f^2}{n^2}$

$T=mg = \frac{4 \mu l^2f^2}{n^2}$  , n = 2

$T = (m-2.2)g = \frac{4 \mu l^2f^2}{n^2}, n = 5$

$\Rightarrow \frac{m}{m-2.2}=\frac{25}{4}$

$\Rightarrow m = 2.619\ kg$

Therefore, μ = 0.002785 kg/ m

Frequency is varied keeping T constant

$T=\frac{4 \mu l^2f^2}{n^2}, f=60 , \ \ n = 2$

$T=\frac{4 \mu l^2f^2}{n^2}, f=? , \ \ n = 7$

$\Rightarrow \frac{60^2}{4}=\frac{f^2}{49}$

f = 210 Hz