Q. 14.7  The motion of a particle executing simple harmonic motion is described by the displacement function,  x(t)=A \; cos(\omega t+\phi ).

 If the initial (t=0)  position of the particle is 1\; cm and its initial velocity is \omega \; cm/s,  what are its amplitude and initial phase angle ? The angular frequency of the particle is  \pi s^{-1}. If instead of the cosine function, we choose the sine function to describe the SHM : x=B\; sin(\omega t+\alpha ),  what are the amplitude and initial phase of the particle with the above initial conditions.

Answers (1)
S Sayak

\omega =\pi\ rad\ s^{-1}

x(t)=Acos(\pi t+\phi )

at t = 0

\\x(0)=Acos(\pi \times 0+\phi )\\ 1=Acos\phi\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (i)

\\v=\frac{\mathrm{d}x(t) }{\mathrm{d} t}\\ v(t)=-A\pi sin(\pi t+ \phi )

at t = 0

\\v(0) =-A\pi sin(\pi \times 0+ \phi ) \\ \omega =-A\pi sin\phi\\\ 1 =-A sin\phi \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)

Squaring and adding equation (i) and (ii) we get

\\1^{2}+1^{2}=(Acos\phi )^{2}+(-Asin\phi )^{2} \\2=A^{2}cos^{2}\phi +A^{2}sin^{2}\phi \\ 2=A^{2}\\ A=\sqrt{2}

Dividing equation (ii) by (i) we get

\\tan\phi =-1\\ \phi =\frac{3\pi }{4},\frac{7\pi }{4},\frac{11\pi }{4}......

x(t)=Bsin(\pi t+\alpha )

at t = 0

\\x(0)=Bsin(\pi \times 0+\alpha )\\ 1=Bsin\alpha \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (iii)

\\v=\frac{\mathrm{d}x(t) }{\mathrm{d} t}\\ v(t)=B\pi cos(\pi t+ \alpha )

at t = 0

\\v(0) =B\pi cos(\pi \times 0+ \alpha ) \\ \omega =B\pi cos\alpha \\\ 1 =B cos\alpha \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (iv)

Squaring and adding equation (iii) and (iv) we get

\\1^{2}+1^{2}=(Bsin\alpha )^{2}+(Bcos\alpha )^{2} \\2=B^{2}sin^{2}\alpha +B^{2}cos^{2}\alpha \\ 2=B^{2}\\ B=\sqrt{2}

Dividing equation (iii) by (iv) we get

\\tan\alpha =1\\ \alpha =\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4}......

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