Physics - SHM Question with Solution | TestHub

Given,$Y=a\sin 2\pi \left(bt-cx\right)$...(i)

and standard form$Y=A\sin \left(\omega t-kx\right)$

Thus, we have

$\omega =2\pi b$and$K=2\pi c$

or$\frac{2\pi }{\lambda }=2\pi c\Rightarrow c=\frac{1}{\lambda }$

$\therefore$Speed of wave,

$v=\frac{\omega }{K}=\frac{2\pi b}{2\pi c}=\frac{b}{c}$

Differentiating Equation (i) w.r.t.t, we get

$\frac{dY}{dt}=$Speed of particle,

$v_p=a\cos 2\pi \left(bt-cx\right)\cdot 2\pi b$

$v_p=2\pi ab\cos 2\pi \left(bt-cx\right)$

For$v_{p_{\mathrm{m}\mathrm{a}\mathrm{x}}}=2\pi ab$

According to the question's condition,

$v_{p_{\mathrm{m}\mathrm{a}\mathrm{x}}}=2v$

$\Rightarrow 2\pi ab=\frac{2\cdot b}{c}$(Putting values)

$\Rightarrow \pi a=\frac{1}{c}$or$ac=\frac{1}{\pi }$