question_answer

Two springs, of force constants\[{{I}_{0}}=\frac{E}{R}=\frac{5}{5}=1\,A\]and\[\tau =\frac{L}{R}=\frac{10}{5}=2\,s\]are connected to a mass m as shown. The frequency of oscillation of the mass is \[f\]. If both\[(\therefore -t/\tau =\frac{-2}{2}=-1)\]and\[J=\frac{i}{\pi {{a}^{2}}}\]are made four times their original values, the frequency of. oscillation becomes       AIEEE  Solved  Paper-2007

A)  \[\oint{B.dl}={{\mu }_{0}}.{{i}_{enclosed}}\]    

B)                                         \[\oint{B.dl}={{\mu }_{0}}.{{i}_{enclosed}}\]                                     

C)         \[x=2\times {{10}^{-2}}\]                                            

D)         \[cos\text{ }\pi t\]

Correct Answer: D

Solution :

The frequencies of oscillation in this situations is given by    \[({{M}_{O}}-17{{M}_{n}}){{c}^{2}}\]      \[p-n\]      (same direction)                 \[hvc\] \[hv/{{c}^{2}}\]\[hv/c\] \[v={{V}_{0}}+gt+f{{t}^{2}}.\]\[x=0\]\[t=0,\]\[(t=1)\] \[{{V}_{0}}+2g+3f\]\[{{V}_{0}}+g/2+f/3\] and\[{{V}_{0}}+g+f\]                                 \[{{V}_{0}}+g/2+f\]