Questions
An object of mass 0.2 kg executes simple harmonic along X-axis with frequency of Hz. At the position x = 0.04 m, the object has kinetic energy of 0.5 J and potential energy of 0.4 J amplitude of oscillation in meter is equal to
detailed solution
Correct option is B
E = 12mω2A2 ⇒ E = 12m(2πf)2A2⇒A = 12πf2EmPutting values we obtain,A = 12π(25π)2×(0.5+0.4)0.2 ⇒ A = 0.06 mTalk 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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