JEE Mains · Physics · STD 11 - 13. oscillations
A particle is subjected two simple harmonic motions as :
\(\mathrm{x}_1=\sqrt{7} \sin 5 \mathrm{tcm}\)
and \(x_2=2 \sqrt{7} \sin \left(5 t+\frac{\pi}{3}\right) \mathrm{cm}\)
where x is displacement and \(t\) is time in seconds. The maximum acceleration of the particle is \(\mathrm{x} \times 10^{-2} \mathrm{~ms}^{-2}\). The value of x is :
- A \(175\)
- B \(25 \sqrt{7}\)
- C \(5 \sqrt{7}\)
- D \(125\)
Answer & Solution
Correct Answer
(A) \(175\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{x}_1=\sqrt{7} \sin 5 \mathrm{t} \\ & \mathrm{x}_2=2 \sqrt{7} \sin \left(5 \mathrm{t}+\frac{\pi}{3}\right)\end{aligned}\) From phasor, \(\therefore\) Amplitude of resultant \(\mathrm{SHM}=7\)…
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