CBSE-XI-Physics
12: Simple Harmonic Motion
- #15Two bodies A and B of equal mass are suspended from two separate massless springs of spring constant k1 and k2 respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is
(a) k1/k2
(b)
k1/k2
(c) k2/k1
(d)
k2/k1digAnsr: dAns : (d)`` \sqrt{\frac{{k}_{2}}{{k}_{1}}}``
Maximum velocity, v = Aω
where A is amplitude and ω is the angular frequency.
Further, ω = `` \sqrt{\frac{k}{m}}``
Let A and B be the amplitudes of particles A and B respectively. As the maximum velocity of particles are equal,
`` i.e.{v}_{A}={v}_{B}``
`` \,\mathrm{\,or\,},``
`` A{\omega }_{A}=B{\omega }_{B}``
`` \Rightarrow A\sqrt{\frac{{k}_{1}}{{m}_{A}}}=B\sqrt{\frac{{k}_{2}}{{m}_{B}}}``
`` \Rightarrow A\sqrt{\frac{{k}_{1}}{m}}=B\sqrt{\frac{{k}_{2}}{m}}({m}_{A}={m}_{B}=m)``
`` ``
`` \Rightarrow \frac{A}{B}=\sqrt{\frac{{k}_{2}}{{k}_{1}}}``
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