Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • An object of mass 'm' is suspended at the end of a mass less wire of length L and area of cross-section A. Young's modules of the maternal of the wire is Y. if the mass is pulled down slightly its freq. of oscillation along the vertical dir

An object of mass 'm' is suspended at the end of a mass less wire of length L and area of cross-section A. Young's modules of the maternal of the wire is Y. if the mass is pulled down slightly its freq. of oscillation along the vertical direction is :
Option: 1 f=\frac{1}{2\pi }\sqrt{\frac{mA}{YL}}
Option: 2 f=\frac{1}{2\pi }\sqrt{\frac{YL}{mA}}
Option: 3 f=\frac{1}{2\pi }\sqrt{\frac{YA}{mL}}
Option: 4 f=\frac{1}{2\pi }\sqrt{\frac{mL}{YA}}

Answers (1)

best_answer

At equilibrium

T=mg ......(i)

Y=\frac{T / A}{l / L}$

\Rightarrow T=\frac{Y A l}{L} \ldots$(ii)

From
(i) and (ii)

m g=\frac{Y A l}{L} \ldots$ (iii)

Now for a small displacement x

\begin{aligned} &\text { Restoring force } &\left.=-\left[T^{\prime}-m g\right]=-\left[\frac{Y A(l+x)}{L}-\frac{Y A l}{L}\right] \right] &=\frac{-Y A x}{L} \end{aligned}Comparing with  F=-m \omega^{2} x

\begin{array}{l} m \omega^{2}=\frac{Y A}{L}\\ \Rightarrow \omega=\sqrt{\frac{Y A}{m L}} \Rightarrow \frac{2 \pi}{T}=\sqrt{\frac{Y A}{m L}} \\ \text { Frequency } =f=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{Y A}{m L}} \end{array}

 

 

 

 

 

Posted by

Deependra Verma

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main BTech CSE cut-offs for students of northeast, UTs drop in 2 years; over 2,000 seats reserved in NITs
anam-khan

BSEB Super 50 performed ‘brilliantly’ in JEE Main, 23 scored above 95 percentile: Bihar Board chairman
anam-khan

IIIT Allahabad cut-offs for BTech ECE, IT, IT-Business Informatics goes up in last 3 years; seat matrix

Latest Question