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JEE Mains · Physics · STD 11 - 13. oscillations

A particle of mass \(m\) is moving along a trajectory given by
\(x = x_0 + a\, cos\,\omega_1 t\)
\(y = y_0 + b\, sin\,\omega_2t\)
The torque, acing on the particle about the origin, at \(t = 0\) is

  1. A \(m{y_0}a\omega _1^2\hat k\)
  2. B \(m\left( { - {x_0}b + {y_0}a} \right)\omega _1^2\hat k\)
  3. C \( - m\left( { - {x_0}b\omega _2^2 + {y_0}a\omega _1^2} \right)\hat k\)
  4. D Zero
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Answer & Solution

Correct Answer

(A) \(m{y_0}a\omega _1^2\hat k\)

Step-by-step Solution

Detailed explanation

\(\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}=\mathrm{m}\left[-\mathrm{a} \omega_{1}^{2} \cos \omega, \mathrm{t} \hat{\mathrm{i}}-\mathrm{b} \omega_{2}^{2} \sin \omega_{2} \mathrm{t} \hat{\mathrm{j}}\right.\)…
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