LIVE Course for free

Rated by 1 million+ students
Get app now
0 votes
20 views
in Physics by (30.0k points)
closed by
A uniform string of length l and mass m is tied at both ends by the tension T, its frequency will be -
1. \(\frac{1}{{2l}}\sqrt {\frac{T}{m}} \)
2. \(\frac{1}{2}\sqrt {\frac{T}{m}} \)
3. \(2l\sqrt {\frac{T}{m}} \)
4. \(\frac{1}{2}\sqrt {\frac{T}{{ml}}} \)

1 Answer

0 votes
by (54.3k points)
selected by
 
Best answer
Correct Answer - Option 4 : \(\frac{1}{2}\sqrt {\frac{T}{{ml}}} \)

Concept:

The frequency of oscillation of a string stretched under tension T is given by 

\(f =\frac{1}{2l}\sqrt {\frac{T}{μ}}{}\)

Where f = frequency of oscillation, n = number of loops, l = Length of string, T =Tension, and μ= linear density

Calculation:

Given, mass of string m

length of string l

linear mass density

\(\mu = \frac{m}{l}\)

\(\implies f =\frac{1}{2l}\sqrt{\frac{T}{(m/l)}}{}\)

\(\implies f =\frac{1}{2l}\sqrt{\frac{Tl}{(m)}}{}\)

\(\implies f =\frac{1}{2}\sqrt{\frac{Tl}{(ml^2)}}{}\)

Writing l = √ (l)2

\(\implies f =\frac{1}{2}\sqrt{\frac{T}{(ml)}}{}\)

So, the correct option is \(\frac{1}{2}\sqrt {\frac{T}{{ml}}} \)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...