Questions
A particle is executing simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm/sec. The distance of the particle from the mean position when its speed becomes 5 cm/s is
detailed solution
Correct option is C
v = ωa2-y2At y = 0, vmax = 10 cm/sAs a = 4 cm, vmax = ωa hence ω = 2.5 rad/sHence, 5 = 2.516-y2or y = 23 cmTalk 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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