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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Fachur

Answered 2021-11-23
Author has **17** answers

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:

${A}^{\prime}=28.3cm$

At equilibrium postion the energy is pure kinetic energy:

Step 2

The maximum speed is given by:

where

Substitution of the previous relation into the the expression for kinetic energy gives:

Step 3

After doubline the kinetic energy

thereby:

substituing the given value

Nancy Johnson

Answered 2021-11-24
Author has **17** answers

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.

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

The total mechanical energy is

1)

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

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

The amplitude is 28.28 cm when E is doubled.

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