Question

A body of mass 10 kg oscillates as per the equation,

Y = 8 sin 100πt +6 cos 100πt (in SI). Its amplitude of oscillation is​

Answer 1

There are one to two methods to solve this ques. I'll use the phaser method.

☆Phaser method:

》In this, first convert the superimposed wave in proper sine equation of wave.

》Then solve the waves in the form of vectors by putting them at axis.

☆Let's solve:

》$\sf Y= 8sin(100 \pi t +0) + 6 sin(100 \pi t + \dfrac{\pi}{2})$

》Now, see the above image.

》phase difference between them is 90°.

》Resultant = (6²+8²)½

》amplitude = 10m

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