Question

The displacement (in m) equation of a particle is x=3 sin 2t+4 cos2t. The amplitude (in m) and maximum velocity (in m/s) will be respectively.

- 5,10
- 4,2
- 3,4
- 3,2

Solution

The correct option is **A** 5,10

Given,

x=3 sin 2t+4 cos 2t

Rewriting the above equation as,

x=3 sin 2t+4 sin (2t+π2) ....(i)

∵sin(π2+θ)=cosθ

Equation (i) represents that displacement (x) of particle is due to superposition of two independent SHM, with a phase difference of π2

⇒x=a1sinωt+a2sin(ωt+ϕ)

From the phases diagram given below:

The resultant amplitude of the particle will be given by,

⇒a=√a21+a22+2a1a2cosϕ

Where amplitude of individual SHM is, a1=3 m,& a2=4 m and phase difference ϕ=π2

⇒a=√a21+a22=√32+42

∴a=5 m

Maximum velocity of the particle is given by,

⇒(vmax)=aω

⇒vmax=5×2=10 m/s

Given,

x=3 sin 2t+4 cos 2t

Rewriting the above equation as,

x=3 sin 2t+4 sin (2t+π2) ....(i)

∵sin(π2+θ)=cosθ

Equation (i) represents that displacement (x) of particle is due to superposition of two independent SHM, with a phase difference of π2

⇒x=a1sinωt+a2sin(ωt+ϕ)

From the phases diagram given below:

The resultant amplitude of the particle will be given by,

⇒a=√a21+a22+2a1a2cosϕ

Where amplitude of individual SHM is, a1=3 m,& a2=4 m and phase difference ϕ=π2

⇒a=√a21+a22=√32+42

∴a=5 m

Maximum velocity of the particle is given by,

⇒(vmax)=aω

⇒vmax=5×2=10 m/s

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