Question

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

Answer 1

Answer:28.28 cm

Explanation:

Given

Amplitude $A=20\ cm$

Total energy of system is equal to the Elastic Potential Energy of the spring mass system at maximum amplitude which is given by

$E=U=\frac{1}{2}kA^2$

$E=\frac{1}{2}kA^2$

When energy is doubled then

$2E=\frac{1}{2}kA'^2$

$2\times \frac{1}{2}kA^2= \frac{1}{2}kA'^2$

$A'=\sqrt{2}A$

$A'=1.414\times 20$

$A'=28.28\ cm$

Answer 2

When the energy doubled, the amplitude become "28.28 cm".

Given:

• Amplitude, A = 20 cm

As we know the relation,

→ $E = U = \frac{1}{2}kA^2$

When the energy is doubled, we get

→ $2E = \frac{1}{2}kA'^2$

→  $A' = \sqrt{2A}$

By substituting the value, we get

        $= 1.414\times 20$

        $= 28.28 \ cm$

Thus the above response is right.

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