Question

Displacement time equation of a particle executing
Simple harmonic motion is x=4sinot+3sin Ot+
Here x is in cm and t in sec. The amplitude of
oscillation of particle is approximately
(1) 7 cm
(2) 5 cm
(3) 6 cm
(4) 9 cm
​

Answer 1

Explanation:

The general equation for simple harmonic motion along the x-axis results from a straightforward application of Newton's second law to a particle of mass m acted on by a force: F = -kx, where x is the displacement from equilibrium and k is called the spring constant.

Answer 2

Answer:

5cm

Explanation:

NOTE: here I am assuming that phase difference is 90 because the expression in your qn is not clear

The reason is that, we can consider this as a form of combining two SHM's whose equations are 4sin(wt) and 3sin(wt)

When such SHM's are superimposed the net amplitude 8s given by

(A^2+A'^2+2AA'cosø)^1/2

So in this case, answer is

(3^2+4^2+2(3)(4)cos(90))=5cm

Hope this answer helped you