A block periodic on a spring has AN amplitude of twenty cm. What's going to the amplitude be if the whole energy is doubled?

In the simple harmonic motion, the energy is conversed 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 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

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Cariglinom5

Answered question

2021-11-07

A block periodic on a spring has AN amplitude of twenty cm. What's going to the amplitude be if the whole energy is doubled?

Answer & Explanation

Greg Snyder

Beginner2021-11-08Added 11 answers

In the simple harmonic motion, the energy is conversed 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

\frac{1}{2} k A^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2}

The total mechanical energy is

E = \frac{1}{2} k A^{2}

A = \sqrt{\frac{2 E}{k}}

As shown by equation (1), the amplitude is directly proportional to root square root of E

A = \sqrt{E}

As k is constant, we can find the amplitude for two instants when E is doubled by

\frac{A_{1}}{A_{2}} = \frac{\sqrt{E}}{\sqrt{2 E}}

\frac{A_{1}}{A_{2}} = \frac{1}{\sqrt{2}}

A_{2} = \sqrt{2} A_{1}

A_{2} = \sqrt{2} (20 c m)

A_{2} = 28.28 c m

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