Question

If y = A sin(wt - kx), then the value of dy/dx is
(A, w and t are constant)
(1) A cos(wt - kx) (2) -Aw cos(wt- kx)
(3) Ak cos(wt - kx) (4) -Ak cos(wt - kx)​

Answer 1

Explanation:

$y =A \: sin(wt - kx) \\ differentiating \: w.r.t. \: x \: on \: both \: \\ sides \: we \: find : \frac{dy}{dx} = \frac{d}{dx} A \: sin(wt - kx) \\ = A \frac{d}{dx} \: sin(wt - kx) \\ = A \: cos(wt - kx)\frac{d}{dx} \: (wt - kx) \\ = A \: cos(wt - kx) \: (0 - k \times 1) \\ = A \: cos(wt - kx) \: ( - k)\\\frac{dy}{dx} = - kA \: cos(wt - kx)$

Answer 2

$\begin{gathered} \tt \: Force \: acting \: on \: mass \: m_1 \\ F = m_1a \\ \implies \: m_1 = \frac{F}{a} = \frac{2}{5} \: kg \\ \tt Force \: acting \: on \: mass \: m_2 \\ F = m_2a \\ \implies \: m_2 = \frac{F}{a} = \frac{2}{7} \: kg \ \\ \tt \: \ now \: when \: masses \: are \: tied \: together \: then \: \\ \rm \: total \: mass = m_1 + m_2 \\ = \frac{2}{5} + \frac{2}{7} \\ = \frac{14 + 10}{35} = \frac{24}{35} \: kg \\ \tt let \: the \: acceleration \: produced \: be \: a \\ \bf \: hence \\ a = \frac{F}{m_1 + m_2} = \frac{2}{ \frac{14}{35} } = \frac{2 \times 35}{14} \\ = \frac{2 \times 7 \times 5}{7 \times 2} = 5 \: m/ {s}^{2} \\ thus \: a= 5 \: m/ {s}^{2} \\ \end{gathered}$