Questions
A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is l0 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is
detailed solution
Correct option is C
vmax = aω ⇒ ω = vmaxa = 104Now, v = ωa2-y2 ⇒ v2 = ω2(a2-y2)⇒y2 = a2-v2ω2⇒Y =A32= 23 cmTalk to our academic expert!
Similar Questions
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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