Question

Two simple harmonic motions are represented by
$y_1=5(\sin 2\pi t+\sqrt{3}\cos 2\pi t)$
$y_2=5(\sin 2\pi t+\frac{\pi }{4})$
The ratio of the amplitude of two S.H.M's is

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $1:\sqrt{3}$

Answer 1

The correct option is C $2:1$

Here, y1=5(sin2πt+3√cos2πt)y1=5(sin⁡2πt+3cos⁡2πt)
y2=5sin(2πt+π4)y2=5sin⁡(2πt+π4)
as of the form of y1=αsin2πt+βcos2πty1=αsin⁡2πt+βcos⁡2πt
Let α=rcosθ=5,β=rsinθ=53√α=rcos⁡θ=5,β=rsin⁡θ=53
∴y1=rcosθsin2πt+rsinθcos2πt=rsin(2πt+θ)∴y1=rcos⁡θsin⁡2πt+rsin⁡θcos⁡2πt=rsin⁡(2πt+θ)
also, α2+β2=r2cos2θ+r2sin2θ=r2α2+β2=r2cos2⁡θ+r2sin2⁡θ=r2
or r=α2+β2−−−−−−√=512+(3√)2−−−−−−−−−√=10r=α2+β2=512+(3)2=10
∴y1=10sin(2πt+θ)∴y1=10sin⁡(2πt+θ)
∴A1A2=105=2