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Calculate total energy of a particle of ...

Calculate total energy of a particle of mass `2.0xx10^(-3)kg` oscillating simple harmonically with angular frequency `"0.45 rad s"^(-1)` and amplitude `1.0xx10^(-3)m`.

Text Solution

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Given, `m=2xx10^(-3)kg, omega="0.45 rad s"^(-1) and A =1xx10^(-3)m`.
We know that total energy `=(1)/(2)n omega^(2)A(2)=(1)/(2)xx2xx10^(-3)xx(0.45)^(2)(10^(-3))^(2)`
i.e. `E=2.025xx10^(-10)J`.
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Knowledge Check

  • The wave number of a wave of light is 2.0xx10^(14)cm^(-1) . The frequency of this light is

    A
    `6.6xx10^(3)s^(-1)`
    B
    `6.6xx10^(-3)e^(-1)`
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    D
    `6.0xx10^(-14)s^(-1)`

  • Unertainty in the position of an electron (mass =9.1xx10^(-31) kg) moving with a velocity 300ms^(-1) accurate upto 0.001% will be h=6.63xx10^(-34)Js

    A
    `5.76xx10^(-2)m`
    B
    `1.92x10^(-2)m`
    C
    `3.84xx10^(-2)m`
    D
    `19.2xx10^(-2)m`

  • Uncertainty in the position of an electron (mass = 9.1 xx 10^(-31) kg) moving with a velocity 300 m s^(-1) , accurate up to 0.001% will be:

    A
    `19.2 xx 10^(-2) m`
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    C
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    D
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