logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
0 votes

Find the time period of S.H.M for the following spring - mass system

$(a)\;2 \pi \sqrt{m\;(\large\frac{1}{K_{1}}+\large\frac{4}{K_{2}}+\large\frac{1}{K_{3}}+\large\frac{4}{K_{4}}+\large\frac{1}{K_{5}})}\qquad(b)\;2 \pi \sqrt{m\;(\large\frac{1}{K_{1}}+\large\frac{1}{K_{2}}+\large\frac{1}{K_{3}}+\large\frac{1}{K_{4}}+\large\frac{1}{K_{5}})}\qquad(c)\;2 \pi \sqrt{m\;(K_{1}+4K_{2}+K_{3}+4K_{4}+K_{5})}\qquad(d)\;None\;of\;these$

Can you answer this question?
 
 

1 Answer

0 votes
Answer : (d) None of these
Explanation :
Let the springs $\;K_{1} , K_{2} , K_{3} , K_{4} ,K_{5}\;$ move by $\;x_{1} ,x_{2} ,x_{3} ,x_{4} ,x_{5}\;.$
Then the mass 'm' moves by
$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x\;(assume)$
As the system is in equilibrium , the tension in the springs be T
By considering the F.B.D's
$2T=K_{4}x_{4} \qquad \; therefore\; , x_{4}=\large\frac{2T}{K_{4}} \;$ similarly for $\;K_{2}\;$ also
and $\;K_{5}x_{5}=T\;,K_{1}x_{1}=T\;,K_{3}x_{3}=T$
$x=\large\frac{T}{K_{1}}+\large\frac{4T}{K_{2}}+\large\frac{T}{K_{3}}+\large\frac{4T}{K_{4}}+\large\frac{T}{K_{5}}$
$x=T\;(\large\frac{1}{K_{1}}+\large\frac{4}{K_{2}}+\large\frac{1}{K_{3}}+\large\frac{4}{K_{4}}+\large\frac{1}{K_{5}})$
Let $\;(\large\frac{1}{K_{1}}+\large\frac{4}{K_{2}}+\large\frac{1}{K_{3}}+\large\frac{4}{K_{4}}+\large\frac{1}{K_{5}})=f$
$x=Tf$
$T=\large\frac{x}{f} \qquad (T=ma)$
$ma=\large\frac{x}{f}$
$a=\large\frac{x}{mf}$
$w=\sqrt{\large\frac{1}{mf}}$
$T=\large\frac{2\pi}{w}=2 \pi \sqrt{mf}$
answered Mar 10, 2014 by yamini.v
edited Mar 10, 2014 by yamini.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...