A block oscillating on a spring has an amplitude of 20 cm. What will be the amplitude if the maximum kinetic energy is doubled?

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Fachur · Accepted Answer

Step 1At equilibrium postion the energy is pure kinetic energy:Ek=12mvmax2Step 2The maximum speed is given by:vmax=ωAwhere ω is angular frequency and A is the amplitude of the spring.Substitution of the previous relation into the the expression for kinetic energy gives:Ek=12mA2ω2Step 3After doubline the kinetic energy Ek′=2⋅Ek, the ratio of the kinetic energies is:Ek′Ek=12mA′2ω212mA2ω2Ek′Ek=A′2A2thereby:2=A′2A2A′=2Asubstituing the given value A=20cm, final result is obtained:A′=28.3cm

Nancy Johnson · Accepted Answer

Step 1  In the simple harmonic motion, the energy is conserved and the total mechanical energy E has an expression in terms of the force constant k and amplitude A as given by equation (15.21) in the form  E=K+U  12kA2=12mv2+12kx2  The total mechanical energy is  E=12kA2  1) A=2Ek  As shown by equation (1), the amplitude is directly proportional to root square root of E  A=E  As k is constant, we can find the amplitude for two instants when E is doubled by  A1A2=E2E  A1A2=12  A2=2A1  A2=2(20cm)  A2=28.28cm  The amplitude is 28.28 cm when E is doubled.

A block oscillating on a spring has an amplitude of 20 cm. What will be the ampl

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