 # A block oscillating on a spring has an amplitude of 20 cm. What will be the ampl Redemitz4s 2021-11-22 Answered
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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Step 1
At equilibrium postion the energy is pure kinetic energy:
${E}_{k}=\frac{1}{2}m{v}_{max}^{2}$
Step 2
The maximum speed is given by:
${v}_{max}=\omega A$
where $\omega$ is angular frequency and A is the amplitude of the spring.
Substitution of the previous relation into the the expression for kinetic energy gives:
${E}_{k}=\frac{1}{2}m{A}^{2}{\omega }^{2}$
Step 3
After doubline the kinetic energy ${E}_{k}^{\prime }=2\cdot {E}_{k}$, the ratio of the kinetic energies is:
$\frac{{E}_{k}^{\prime }}{{E}_{k}}=\frac{\frac{1}{2}m{A}^{{}^{\prime }2}{\omega }^{2}}{\frac{1}{2}m{A}^{2}{\omega }^{2}}$
$\frac{{E}_{k}^{\prime }}{{E}_{k}}=\frac{{A}^{{}^{\prime }2}}{{A}^{2}}$
thereby:
$2=\frac{{A}^{{}^{\prime }2}}{{A}^{2}}$
${A}^{\prime }=\sqrt{2}A$
substituing the given value $A=20cm$, final result is obtained:
${A}^{\prime }=28.3cm$
###### Not exactly what you’re looking for? Nancy Johnson
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
$E=K+U$
$\frac{1}{2}k{A}^{2}=\frac{1}{2}m{v}^{2}+\frac{1}{2}k{x}^{2}$
The total mechanical energy is
$E=\frac{1}{2}k{A}^{2}$
1) $A=\sqrt{\frac{2E}{k}}$
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
$\frac{{A}_{1}}{{A}_{2}}=\frac{\sqrt{E}}{\sqrt{2E}}$
$\frac{{A}_{1}}{{A}_{2}}=\frac{1}{\sqrt{2}}$
${A}_{2}=\sqrt{2}{A}_{1}$
${A}_{2}=\sqrt{2}\left(20cm\right)$
${A}_{2}=28.28cm$
The amplitude is 28.28 cm when E is doubled.