Question

A particle is executing shm about y=0 along y axis. its position at an instant is given by y= 5(sin3*3.14*t + cos 3*3.14t) . the amplitude of oscillation is

Answer 1

equation of particle in SHM : y = 5( sin3*3.14t + cos3*3.14t)
in general form we can again write it y = 5sin3πt + 5cos3πt [ π ≈ 3.14 ]

Now, we should use mathematical concept .
If f(x) =a sinx +b cosx ,
= $\bold{\sqrt{a^2+b^2}[\frac{a}{\sqrt{a^2+b^2}}sinx+\frac{b}{\sqrt{a^2+b^2}}cosx]}$
Let $\bold{\frac{a}{\sqrt{a^2+b^2}}=cos{\alpha}}\\\\\bold{\frac{b}{\sqrt{a^2+b^2}}=sin{\alpha}}$
then, f(x) = $\bold{\sqrt{a^2+b^2}sin(x+\alpha)}}$

Use this application here,
Then, y = 5sin3πt + 5cos3πt
y = $\bold{\sqrt{5^2+5^2}sin(3\pi t + \alpha)}}$
= 5√2sin(3πt + α)
here α = tan⁻¹ (5/5) = π/4

Now, compare this equation to standard equation of SHM
y = Asin(ωt + kx)
we get A = 5√2

Hence, amplitude of oscillation = 5√2 unit

Answer 2

5(sin3πt+cos3πt)

simply open the bracket

5sin3πt+5cos3πt

since phase difference between sin and cos is 90°

by using vector concept

A=√5^2+5^2

A=√50 or 5√2