question_answer

The frequency of oscillator of the springs as shown in figure will be:

A) \[\frac{1}{2\pi }\sqrt{\frac{({{k}_{1}}+{{k}_{2}})m}{{{k}_{1}}{{k}_{2}}}}\]               

B) \[\frac{1}{2\pi }\sqrt{\frac{{{k}_{1}}{{k}_{2}}}{({{k}_{1}}+{{k}_{2}})m}}\]

C)                        \[\frac{1}{2\pi }\sqrt{\frac{k}{m}}\]        

D)        \[2\pi \sqrt{\frac{k}{m}}\]

Correct Answer: B

Solution :

                Let at any instant the displacement of the body m from the equilibrium position is y. As force constants of both the springs are different the extensions in their lengths will also be different. Let extension in length of spring \[{{0}^{o}}C\] be \[({{T}^{o}}C)\] and that of \[E=AT-\frac{1}{2}B{{T}^{2}}\] be \[A=16,\,B=0.08\].Then \[\therefore \] Force exerted by spring \[E=16T-\frac{1}{2}\times 0.08\times {{T}^{2}}\] is \[0=16T-0.04{{T}^{2}}\]                                   ?...(1) Force exerted by spring \[{{k}_{2}}\] is \[\Rightarrow \]                                  ?...(2) Multiplying Eq. (1) by \[T=\frac{16}{0.04}={{400}^{o}}C\] and Eq. (2) by \[\frac{A}{2}\] and Adding \[\therefore \] As both the springs are in series they will apply same force on the body that is \[C={{C}_{1}}+{{C}_{2}}\] Then \[=\frac{{{K}_{1}}{{\varepsilon }_{0}}A/2}{d}+\frac{{{K}_{2}}{{\varepsilon }_{0}}A/2}{d}=\frac{({{K}_{1}}+{{K}_{2}}){{\varepsilon }_{0}}A}{d}\] \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi Ni}{a}\]   \[N=25,\text{ }D=10\text{ }cm=10\times {{10}^{-2}}\text{ }m\] Hence, effective force constant is \[a=\frac{D}{2}=\frac{10\times {{10}^{-2}}}{2}m=0.05m\] So, frequency is \[i=4A\] Alternative: As springs are connected in series, effective force constant \[\therefore \] \[B={{10}^{-7}}\times \frac{2\times 3.14\times 25\times 4}{0.05}\] Hence, frequency of oscillation is \[B=1.256\times {{10}^{-3}}T\] \[|m|=\frac{{{f}_{0}}}{{{f}_{e}}}\left( 1+\frac{{{f}_{e}}}{D} \right)\]