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JEE Main 2003 (AIEEE) · Spring Systems

Spring Systems question from JEE Main 2003 (AIEEE)

Q95JEE Main 2003 · AIEEEEasy

Spring Systems

A mass MM is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHMSHM of time period T.T. If the mass is increased by m.m. the time period becomes 5T3\frac{5T}{3}. Then the ratio of mM\frac{m}{M} is

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Answer: (c)

  1. T=2πMkT = 2\pi\sqrt{\frac{M}{k}}, so (T′T)2=M+mM\left(\frac{T'}{T}\right)^2 = \frac{M+m}{M}.
  2. (53)2=259=1+mM\left(\frac{5}{3}\right)^2 = \frac{25}{9} = 1 + \frac{m}{M}
  3. mM=259−1=169\frac{m}{M} = \frac{25}{9} - 1 = \frac{16}{9}
  4. So the answer is (c) 169\frac{16}{9}.
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