Question

A transverse wave on a string is described by the equation
$y(x,t)=\left(2.20cm\right)\sin \left[\right(130rad/s)t+(15rad/m\left)x\right]$, then find the approximate maximum transverse acceleration of a point on the string.

• A. $300{\text{m/s}}^2$
• B. $372{\text{m/s}}^2$
• C. $410{\text{m/s}}^2$
• D. $450{\text{m/s}}^2$

Answer 1

Answer: (B)

$372{\text{m/s}}^2$

$y(x,t)=2.20cm\sin [\left(130rad/s\right)t+\left(15rad/m\right)x]$

Differentiate wrt ${}^{\prime }{t}^{\prime }$, keeping $x$ constant

${\left(\frac{\partial y}{\partial t}\right)}_x=\left(2.20cm\right)\cos [\left(130rad/s\right)t+\left(15rad/m\right)x]\cdot \left(130rad/s\right)$

Again differentiating ${\left(\frac{\partial y}{\partial t}\right)}_x$ wrt ${}^{\prime }{t}^{\prime }$

$$\left( \dfrac { \partial ^{ 2 }y }{ \partial t^{ 2 } } \right) _{ x }=\left| (2.20cm)-\sin { \left[ \left( 130rad/s \right) t+\left( 15rad/m \right) x \right] } \cdot \left( 130rad/s \right) \left( 130rad/s \right) \right|

Now, the maximum transverse acceleration can be found by maximizing ${\left(\frac{{\partial }^{2}y}{\partial t^2}\right)}_x$ for that $\sin [\left(130rad/s\right)t+\left(15rad/m\right)x]=1$

$\therefore {\left(\frac{{\partial }^{2}y}{\partial t^2}\right)}_x=\left(2.20cm\right)\times \left(16900{rad}^2/s^2\right)=37180cm/s^2=371.80m/s^2$

Approximate max. transverse acceleration $=372m/s^2$