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. |
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