A string of length 1,m and linear mass densit | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The wave speed $v$ is given by $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.01}} = 100\,m/s$.For a string fixed at both ends,the frequencies are

Vedclass · Accepted Answer

$n_2 = \frac{f_1 + f_2}{2} = 75\,Hz$

Vedclass · Answer

(D) The wave speed $v$ is given by $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{0.01}} = 100\,m/s$.For a string fixed at both ends,the frequencies are $f_n = \frac{nv}{2L} = \frac{n 	imes 100}{2 	imes 1} = 50n\,Hz$. Thus,$f_1 = 50\,Hz, f_2 = 100\,Hz, f_3 = 150\,Hz$.For a string fixed at one end,the frequencies are $n_k = \frac{(2k-1)v}{4L} = \frac{(2k-1) 	imes 100}{4 	imes 1} = 25(2k-1)\,Hz$ for $k=1, 2, 3, \dots$. Thus,$n_1 = 25\,Hz, n_2 = 75\,Hz, n_3 = 125\,Hz$.Comparing these,$n_2 = \frac{f_1 + f_2}{2} = \frac{50 + 100}{2} = 75\,Hz$.

Similar Questions

$A$ wave travels on a light string. The equation of the wave is $Y = A \sin (kx - \omega t + 30^o)$. It is reflected from a heavy string tied to an end of the light string at $x = 0$. If $64\%$ of the incident energy is reflected,the equation of the reflected wave is:

$A$ stretched string fixed at both ends has $m$ nodes,then the length of the string will be

At nodes in stationary waves,

Stationary waves can be produced in

Vedclass Test Series

Exam Paper Generator

Online Exam Module