Question

if a particle performing shm of amplitude 10 cm at what distance from the mean position is the kinetic energy 2 time the potential energy​

Answer 1

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Given:

Amplitude
`\displaystyle\sf A = 10\,\text{cm} = 0.1\,\text{m}`

We want to find
`\displaystyle\sf x` such that the kinetic energy
`\displaystyle\sf KE` is twice the potential energy
`\displaystyle\sf PE`.

Using the formulas for kinetic energy and potential energy in SHM:

`\displaystyle\sf KE = (1)/(2)mv^(2)`

`\displaystyle\sf PE = (1)/(2)kx^(2)`

Since we are considering the same particle, the mass
`\displaystyle\sf m` cancels out. Therefore, we can compare the expressions for
`\displaystyle\sf KE` and
`\displaystyle\sf PE`:

`\displaystyle\sf (1)/(2)kx^(2) = 2\left((1)/(2)kA^(2)\right)`

Simplifying this equation:

`\displaystyle\sf kx^(2) = 2kA^(2)`

Dividing both sides by
`\displaystyle\sf k`:

`\displaystyle\sf x^(2) = 2A^(2)`

Taking the square root of both sides:

`\displaystyle\sf x = √(2)A`

Substituting the value of
`\displaystyle\sf A`:

`\displaystyle\sf x = √(2) * 0.1\,\text{m}`

Calculating the value:

`\displaystyle\sf x = 0.1414\,\text{m}`

Converting to centimeters:

`\displaystyle\sf x = 0.1414 * 100\,\text{cm}`

`\displaystyle\sf x \approx 14.14\,\text{cm}`

Therefore, the distance from the mean position at which the kinetic energy is twice the potential energy is approximately
`\displaystyle\sf 14.14\,\text{cm}`.

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Answer 2

The equation for kinetic energy of shm system is given by;

KE=0.5mw²(A²-x²)

where m is the mass of the object

w² is the angular frequency

A is the amplitude

and x is the positive

The equation for the potential energy is given by;

PE=0.5mw²x²

Solve KE=2PE and you get ur answer