Question

The motion of a particle is given by x = A sin ωt + B cos ωt. The motion of the particle is

विकल्प

• not simple harmonic

• simple harmonic with amplitude A + B

• simple harmonic with amplitude (A + B)/2

• simple harmonic with amplitude

Answer 1

simple harmonic with amplitude \[\sqrt{A^2 + B^2}\]

x = A sin ωt + B cos ωt      ...(1)

\[\text { Acceleration }, \]

\[ a = \frac{\text {d}^2 x}{\text {dt}^2} = \frac{\text {d}^2}{\text{dt}^2}(\text {A}\sin\omega t + \text {B} \cos \omega  t)\]

\[ = \frac{\text{d}}{\text {dt}}(\text { A }\omega \cos \omega t - \text { B }\omega \sin \omega t) \]

\[ = - \text { A } \omega^2 \text { sin }\omega t - \text { B }\omega^2 \cos \omega t \]

\[ = - \omega^2 (\text { A }\sin \omega t + \text { B }\cos \omega t )\]

\[ = - \omega^2 x\]

For a body to undergo simple harmonic motion,
acceleration, a =\[-\] kx.     ...(2)

Therefore, from the equations (1) and (2), it can be seen that the given body undergoes simple harmonic motion with amplitude,  A

\[= \sqrt{A^2 + B^2}\]