The displacement of a particle is represented by the equation y= sin 3 ω t . The motion is

Question

The displacement of a particle is represented by the equation y= sin 3 ω t . The motion is (A) non-periodic (B) periodic but not simple harmonic (C) simple harmonic with period 2 π / ω (D) simple harmonic with period π / ω

Tardigrade Admin · Accepted Answer

Given, y= sin 3 ω t=(1/4)[3 sin ω t- sin 3 ω t] As this motion is not represented by single harmonic function, hence it is not SHM. As this motion involves sine and cosine functions, hence it is periodic motion.

Q. The displacement of a particle is represented by the equation $y = sin^{3} ω t .$ The motion is

non-periodic

periodic but not simple harmonic

simple harmonic with period $2 π / ω$

simple harmonic with period $π / ω$

Given, $y = sin^{3} ω t = \frac{1}{4} [3 sin ω t - sin 3 ω t]$ As this motion is not represented by single harmonic function, hence it is not SHM. As this motion involves sine and cosine functions, hence it is periodic motion.

Q. The displacement of a particle is represented by the equation y=sin3ωt. The motion is

non-periodic

periodic but not simple harmonic

simple harmonic with period 2π/ω

simple harmonic with period π/ω

Given, y=sin3ωt=41​[3sinωt−sin3ωt] As this motion is not represented by single harmonic function, hence it is not SHM. As this motion involves sine and cosine functions, hence it is periodic motion.

Q. The displacement of a particle is represented by the equation $y = sin^{3} ω t .$ The motion is

simple harmonic with period $2 π / ω$

simple harmonic with period $π / ω$

Given, $y = sin^{3} ω t = \frac{1}{4} [3 sin ω t - sin 3 ω t]$ As this motion is not represented by single harmonic function, hence it is not SHM. As this motion involves sine and cosine functions, hence it is periodic motion.