JEE MAIN - Physics (2021 - 20th July Evening Shift - No. 9)
A particle is making simple harmonic motion along the X-axis. If at a distances x1 and x2 from the mean position the velocities of the particle are v1 and v2 respectively. The time period of its oscillation is given as :
$$T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 - v_2^2}}} $$
$$T = 2\pi \sqrt {{{x_2^2 + x_1^2} \over {v_1^2 + v_2^2}}} $$
$$T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 + v_2^2}}} $$
$$T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 - v_2^2}}} $$
Explanation
$${v^2} = {\omega ^2}({A^2} - {x^2})$$
$${A^2} = x_1^2 + {{v_1^2} \over {{\omega ^2}}} = x_2^2 + {{v_2^2} \over {{\omega ^2}}}$$
$${\omega ^2} = {{v_2^2 - v_1^2} \over {x_1^2 - x_2^2}}$$
$$T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 - v_2^2}}} $$
$${A^2} = x_1^2 + {{v_1^2} \over {{\omega ^2}}} = x_2^2 + {{v_2^2} \over {{\omega ^2}}}$$
$${\omega ^2} = {{v_2^2 - v_1^2} \over {x_1^2 - x_2^2}}$$
$$T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 - v_2^2}}} $$
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