question_answer

A mass is suspended separately by two springs of spring constants \[{{k}_{1}}\] and \[{{k}_{2}}\] in successive order. The time periods of oscillations in the two cases are \[{{T}_{1}}\] and \[{{T}_{2}}\] respectively. If the same mass be suspended by connecting the two springs in parallel, (as shown in figure) then   the   time   period of oscillations is T. The correct relation is:  [AIPMT 2002]

A)        \[{{T}^{2}}=T_{1}^{2}+T_{2}^{2}\]                    

B)  \[{{T}^{-2}}=T_{1}^{-2}+T_{2}^{-2}\]

C)              \[{{T}^{-1}}=T_{1}^{-1}+T_{2}^{-1}\]

D)       \[T={{T}_{1}}+{{T}_{2}}\]

Correct Answer: B

Solution :

Two simple harmonic motions be written as

\[x=A\sin (\omega t+\delta )\]                  (i)

and       \[y=A\,\,\sin \,\left( \omega t+\delta +\frac{\pi }{2} \right)\]

or          \[y=A\cos (\omega t+\delta )\]                (ii)

Squaring and adding Eqs. (i) and (ii) we obtain

\[{{x}^{2}}+{{y}^{2}}={{A}^{2}}\,[\sin (\omega t+\delta )+{{\cos }^{2}}(\omega t+\delta )]\]

or         \[{{x}^{2}}+{{y}^{2}}={{A}^{2}}\]          \[(\because {{\sin }^{2}}\,\theta +{{\cos }^{2}}\,\theta =1)\]

This is the equation of a circle.

At         \[\,\omega t+\delta =0\,;\,x=0,\,y=A\]

At         \[\omega t+\delta =\frac{\pi }{2}\,;\,x=A,\,y=0\]

At         \[\omega t+\delta =\pi \,\,\,;\,x=0,\,y=-A\]

At         \[\omega t+\delta =\frac{3\pi }{2}\,\,\,;\,x=-A,\,y=0\]

At         \[\omega t+\delta =2\pi \,\,\,;\,x=0,\,y=A\]

Thus, it is obvious that motion of particle is traversed in clockwise direction.