ExamBro
ExamBro
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 :

  1. A \(175\)
  2. B \(25 \sqrt{7}\)
  3. C \(5 \sqrt{7}\)
  4. D \(125\)
Verified Solution

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 \(\mathrm{SHM}=7\)…
From JEE Mains
Explore more questions on app