JEE MAIN - Physics (2023 - 13th April Morning Shift - No. 25)
At a given point of time the value of displacement of a simple harmonic oscillator is given as $$\mathrm{y}=\mathrm{A} \cos \left(30^{\circ}\right)$$.
If amplitude is $$40 \mathrm{~cm}$$ and kinetic energy at that time is $$200 \mathrm{~J}$$, the value of force constant is $$1.0 \times 10^{x} ~\mathrm{Nm}^{-1}$$. The value of $$x$$ is ____________.
Answer
4
Explanation
Given the general equation for displacement in a simple harmonic oscillator:
$$x = A \sin(\omega t + \phi)$$
At the given time, we have:
$$\omega t + \phi = 30^\circ$$
Given the amplitude $$A = 40 \,\text{cm}$$ and the displacement $$x = 40 \times \frac{\sqrt{3}}{2} \,\text{cm} = 20\sqrt{3} \,\text{cm}$$, we can write the kinetic energy, $$KE$$, as:
$$KE = \frac{1}{2}k(A^2 - x^2) = 200$$
Now, substitute the values for $$A$$ and $$x$$:
$$200 = \frac{1}{2}k\left(\frac{1600 - 1200}{100 \times 100}\right)$$
Simplify the equation:
$$400 \times 100 \times 100 = k \times 400$$
Solve for the force constant, $$k$$:
$$k = 10^4 \,\text{Nm}^{-1}$$
Given that the force constant is expressed as $$k = 1.0 \times 10^x \,\text{Nm}^{-1}$$, comparing the values, we get:
$$1.0 \times 10^x = 10^4$$
Thus, the value of $$x$$ is $$4$$.
$$x = A \sin(\omega t + \phi)$$
At the given time, we have:
$$\omega t + \phi = 30^\circ$$
Given the amplitude $$A = 40 \,\text{cm}$$ and the displacement $$x = 40 \times \frac{\sqrt{3}}{2} \,\text{cm} = 20\sqrt{3} \,\text{cm}$$, we can write the kinetic energy, $$KE$$, as:
$$KE = \frac{1}{2}k(A^2 - x^2) = 200$$
Now, substitute the values for $$A$$ and $$x$$:
$$200 = \frac{1}{2}k\left(\frac{1600 - 1200}{100 \times 100}\right)$$
Simplify the equation:
$$400 \times 100 \times 100 = k \times 400$$
Solve for the force constant, $$k$$:
$$k = 10^4 \,\text{Nm}^{-1}$$
Given that the force constant is expressed as $$k = 1.0 \times 10^x \,\text{Nm}^{-1}$$, comparing the values, we get:
$$1.0 \times 10^x = 10^4$$
Thus, the value of $$x$$ is $$4$$.
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