A particle is subjected to two simple harmonic motions as: x_1 = sqrt7sin 5tmathrm~cm and x_2 = 2sqrt7sinleft(5t + fracpi3 ight)mathrm~cm where x is displacement and t is time in seconds. The maximum acceleration of the particle is x times 10^-2mathrm~ms^-2. The value of x is:

Solution & Explanation

### Related Formula A = sqrtA_1^2 + A_2^2 + 2 A_1 A_2 cosphi a_max = omega^2 A ### Core Logic Both SHMs share the same frequency omega = 5mathrm~rad/s with phase difference phi = fracpi3. Amplitudes: A_1 = sqrt7mathrm~cm, A_2 = 2sqrt7mathrm~cm. The combined amplitude is: A = sqrt(sqrt7)^2 + (2sqrt7)^2 + 2(sqrt7)(2sqrt7)cosleft(fracpi3 ight) A = sqrt7 + 28 + 2(7)(2)left(frac12 ight) = sqrt35 + 14 = sqrt49 = 7mathrm~cm Converting to meters: A = 0.07mathrm~m The maximum acceleration is: a_max = omega^2 A = (5)^2 times 0.07 = 25 times 0.07 = 1.75mathrm~ms^-2 Equating to x times 10^-2mathrm~ms^-2: 1.75 = x times 10^-2 implies x = 175 ### Step 1: Final Conclusion The value of x is 175. ### Pattern Recognition When superposing two SHMs of identical frequency, use the standard phasor addition formula to obtain the combined amplitude A. Then compute maximum velocity v_max = omega A or maximum acceleration a_max = omega^2 A directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Simple Harmonic Motion
Phasor addition of SHM amplitudes
Phasor addition of SHM amplitudes

Reference Study Guides

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)