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A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes simple harmonic oscillations with a time period T. If the mass is increased by m then the time period becomes(). The ratio of is
detailed solution
Correct option is A
T = 2πmK ⇒m ∝T2 ⇒ m2m1 = T22T21⇒ M+mM = (54TT)2 ⇒ mM = 916Talk to our academic expert!
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The identical springs of constant ‘K’ are connected in series and parallel as shown in figure A mass M is suspended from them. The ratio of their frequencies of vertical oscillations will be
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