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QJEEM22Ph8274: JEE Main 2022 Physics - Class 11 JEEM22Ph8274 Solution: In figure , mass ' is fixed on mass ' ' which is attached to…

JEE Main 2022 (Online) 25th July Morning Shift Physics - Simple Harmonic Motion Question 53 English

In figure (A)(\mathrm{A}) , mass ' 2 m’ 2 \mathrm{~m}^{\text {' }} is fixed on mass ' m\mathrm{m} ' which is attached to two springs of spring constant k\mathrm{k} . In figure (B), mass ' m\mathrm{m} ' is attached to two springs of spring constant ' k\mathrm{k} ' and ' 2k′2 \mathrm{k}^{\prime} . If mass ' m\mathrm{m} ' in (A) and in (B) are displaced by distance ' x′x^{\prime} horizontally and then released, then time period T1\mathrm{T}_{1} and T2\mathrm{T}_{2} corresponding to (A)(\mathrm{A}) and (B) respectively follow the relation.

Multiple Choice Options

  • A T1 T2=32\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{3}{\sqrt{2}}
  • B T1 T2=32\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{3}{2}}
  • C T1 T2=23\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\sqrt{\frac{2}{3}}
  • D T1T2=23\frac{T_{1}}{T_{2}}=\frac{\sqrt{2}}{3}

Step-by-Step Verified Solution

Both the springs are in parallel combination in both the diagrams so

T1=2π3m2k{T_1} = 2\pi \sqrt {{{3m} \over {2k}}}

and T2=2πm3k{T_2} = 2\pi \sqrt {{m \over {3k}}}

So, T1T2=32{{{T_1}} \over {{T_2}}} = {3 \over {\sqrt 2 }}

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