Question

You are told that at the known positions x1 and x2 ,an oscillating mass m has speeds v1 and v2 respectively. What are the amplitude and angular frequency of the oscillations? (Hint: x(t) = B1cos(wt) + B2sin(wt))

Answer 1

X₁=B₁Cos(Wt₁)+B₂Sin(Wt₁)--------------------1

X₂=B₁Cos(Wt₂)+B₂sin(Wt₂)--------------------2

Differentiating With respect to t

V₁=dX₁/dt =-B₁WSin(Wt₁)+B₂WCos(Wt₁)----------------3

V₂ =dX₂/dt =-B₁WSin(Wt₂)+B₂WCos(Wt₂)----------------4

(W*1)²+3²

X₁²W²+V₁²=W²*(B₁Cos(Wt₁)+B₂Sin(Wt₁))²+(-B₁WSin(Wt₁)+B₂WCos(Wt₁))²

X₁²W²+V₁²=B₁²W²+B₂²W²--------------5

(W*2)²+4²

X₂²W²+V₂²=[W^*(B₁Cos(Wt₂)+B₂sin(Wt₂))]²+[-B₁WSin(Wt₂)+B₂WCos(Wt₂)]²

X₂²W²+V₂²=B₁²W²+B₂²W²-----------------------6

5-6

X₁²W²+V₁²-X₂²W²-V₂²=0

(X₁²-X₂²)W² =(V₂²-V₁²)

so angular freqeuncy

$\omega =\sqrt{\frac{V_{2}^{2}-V_{1}^{2}}{x_{1}^{2}-x_{2}^{2}}}$

substituting W in 5 we get

$X_{1}^{2}\left ( \sqrt{\frac{V_{2}^{2}-V_{1}^{2}}{x_{1}^{2}-x_{2}^{2}}} \right )^{2}+V_{1}^{2} = (B_{1}^{2}+B_{2}^{2}) \times \left ( \sqrt{\frac{V_{2}^{2}-V_{1}^{2}}{x_{1}^{2}-x_{2}^{2}}} \right )^{2}$

So amplitude

$A=B_{1}^{2}+B_{2}^{2}=\sqrt{\frac{x_{1}^{2}V_{2}^{2}-x_{2}^{2}V_{1}^{2}}{V_{2}^{2}-V_{1}^{2}}}$