
When the displacement is half of the amplitude, the ratio of potential energy to the total energy is
A) $\dfrac{1}{2}$
B) $\dfrac{1}{4}$
C) $1$
D) $\dfrac{1}{8}$
Answer
163.2k+ views
Hint:
Here in this question, It is given that when the displacement is half to their amplitude. Then we have to find the ratio between the Potential energy and the Total energy. As for which we can simply write the formula of both and compare them on the basis of their given changes in displacement and amplitude after which we can easily justify the ratio of Potential energy and Total energy.
Formula used :
As we start the solution from the Potential Energy,
${U_p} = \dfrac{1}{2}m{\omega ^2}{y^2}$
Complete step by step solution:
Here we know that, y is the displacement so in the question there is given the value of displacement, so the value of y is,
$y = \dfrac{A}{2}$
By putting the value of y in the formula of Potential energy we get,
${U_p} = \dfrac{1}{2}m{\omega ^2}{\dfrac{A}{4}^2}$
Here in the above equation, we get the value of potential energy now for total energy we know that,
$E = \dfrac{1}{2}m{\omega ^2}{A^2}$
By taking the required ratio for both Potential Energy with Total Energy we get,
$\dfrac{{{U_p}}}{E} = \dfrac{{\dfrac{1}{2}m{\omega ^2}{{\dfrac{A}{4}}^2}}}{{\dfrac{1}{2}m{\omega ^2}{A^2}}}$
By further solving the above equation, we get the requires ratio as,
$\dfrac{{{U_p}}}{E} = \dfrac{1}{4}$
Therefore, the correct answer for the ratio of P.E. with T.E. is $\dfrac{1}{4}$ .
Hence, the correct option is (B).
Hence the correct answer is Option(b).
Note:
For such types of questions one should be aware of all the formulas used and should go through all the concepts used in this question thoroughly.
Here in this question, It is given that when the displacement is half to their amplitude. Then we have to find the ratio between the Potential energy and the Total energy. As for which we can simply write the formula of both and compare them on the basis of their given changes in displacement and amplitude after which we can easily justify the ratio of Potential energy and Total energy.
Formula used :
As we start the solution from the Potential Energy,
${U_p} = \dfrac{1}{2}m{\omega ^2}{y^2}$
Complete step by step solution:
Here we know that, y is the displacement so in the question there is given the value of displacement, so the value of y is,
$y = \dfrac{A}{2}$
By putting the value of y in the formula of Potential energy we get,
${U_p} = \dfrac{1}{2}m{\omega ^2}{\dfrac{A}{4}^2}$
Here in the above equation, we get the value of potential energy now for total energy we know that,
$E = \dfrac{1}{2}m{\omega ^2}{A^2}$
By taking the required ratio for both Potential Energy with Total Energy we get,
$\dfrac{{{U_p}}}{E} = \dfrac{{\dfrac{1}{2}m{\omega ^2}{{\dfrac{A}{4}}^2}}}{{\dfrac{1}{2}m{\omega ^2}{A^2}}}$
By further solving the above equation, we get the requires ratio as,
$\dfrac{{{U_p}}}{E} = \dfrac{1}{4}$
Therefore, the correct answer for the ratio of P.E. with T.E. is $\dfrac{1}{4}$ .
Hence, the correct option is (B).
Hence the correct answer is Option(b).
Note:
For such types of questions one should be aware of all the formulas used and should go through all the concepts used in this question thoroughly.
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