Question

 For any SHM, amplitude is $6cm$ . If instantaneous potential energy if half the total energy then distance of particle from its mean position is
A) $3cm$
B) $4.2cm$
C) $5.8cm$
D) $6cm$

Answer 1

Hint:
Here in this question, it is given that for any simple harmonic motion, the amplitude of that is $6cm$ From which, if the instantaneous potential energy is half to the total energy then we have to find the distance of the particle from its mean position. We can solve it by using the relation between potential energy and displacement of the particle doing SHM

Formula used :
As we know that, the potential energy in the simple harmonic motion is given as,
$U = \dfrac{1}{2}k{y^2}$


Complete step by step solution:
In the above formula of potential energy, the terms stand for,
$k$ is the force constant in the above formula
And $y$ is the distance of the particle from its mean position.
The simple harmonic motion's total energy is given by,
$E = \dfrac{1}{2}k{a^2}$
Similarly, in the above equation term $a$ stands for amplitude in the simple harmonic motion.
As given in the question that,
$U = \dfrac{E}{2}$
As by putting all the values in above equation, we get that,
$\dfrac{1}{2}k{y^2} = \left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{2}} \right)k{a^2}$
By doing further solution we get that,
${y^2} = \dfrac{{{a^2}}}{2}$
Now, putting $a$ ‘s value from the question we get the result as,
${y^2} = \dfrac{{{6^2}}}{2}$
By solution we get, the final as,
$y = \sqrt {18} = 4.2cm$
Therefore, the correct answer is $4.2cm$ .
Hence, the correct option is (B).

Hence the correct answer is Option(B).






Note:
 Relations of potential energy as per with SHM are, Potential energy is therefore simply proportional to the displacement square. As a result, a simple harmonic oscillator's potential energy is equal to one by four times the total energy when the particle is halfway to its end point.