If the length of a stretched string is shorte | vedclass

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

Question

(A) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$, where $l$ is the length, $T$ is the tension, an

Vedclass · Accepted Answer

$2: 1$

Vedclass · Answer

(A) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$, where $l$ is the length, $T$ is the tension, and $m$ is the mass per unit length.Initial frequency: $n_1 = \frac{1}{2l} \sqrt{\frac{T}{m}}$.New length: $l' = l - 0.40l = 0.6l$.New tension: $T' = T + 0.44T = 1.44T$.New frequency: $n_2 = \frac{1}{2l'} \sqrt{\frac{T'}{m}} = \frac{1}{2(0.6l)} \sqrt{\frac{1.44T}{m}}$.Taking the ratio: $\frac{n_2}{n_1} = \frac{\frac{1}{2(0.6l)} \sqrt{\frac{1.44T}{m}}}{\frac{1}{2l} \sqrt{\frac{T}{m}}} = \frac{l}{0.6l} 	imes \sqrt{\frac{1.44T}{T}} = \frac{1}{0.6} 	imes \sqrt{1.44} = \frac{1.2}{0.6} = 2$.Therefore, the ratio of the final to initial frequency is $2: 1$.

If the length of a stretched string is shortened by $40 \%$ and the tension is increased by $44 \%$, then the ratio of the final and initial fundamental frequencies is :

Similar Questions

The fundamental frequency of a sonometer wire is $50 \ Hz$ for a given length and tension. If the length is increased by $25 \%$ while keeping the tension constant,what is the percentage change in the frequency of the second harmonic?

$A$ string is clamped at both ends and it is vibrating in its $4^{th}$ harmonic. The equation of the stationary wave is $Y = 0.3 \sin(0.157 x) \cos(200\pi t)$. The length of the string is ..... $m$ (all quantities are in $SI$ units).

Standing waves are generated on a sonometer string loaded with a cylindrical body. If the cylinder is completely immersed in water,the length of the loops changes by a factor of $2.2$. The specific gravity of the material of the cylinder is

$A$ string in a musical instrument is $50 \ cm$ long and its fundamental frequency is $800 \ Hz$. If a frequency of $1000 \ Hz$ is to be produced,then the required length of the string is ..... $cm$.

If the fundamental frequency of a string is $220 \, cps$,the frequency of the fifth harmonic will be ......... $cps$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module