Question

In the equation $x=A\sin (\omega t+\frac{\pi }{4})$. Match the following, when $x=\frac{A}{2}$

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{(A) Kinetic energy}& \text{(p) Half the maximum value}\\ \text{(B) Potential energy}& \text{(q) 3/4 times the maximum value}\\ \text{(C) Acceleration}& \text{(r) 1/4 times the maximum value}\\ & \text{(s) Cannot say anything}\end{array}$

• A. $A\to q,B\to p,C\to s$
• B. $A\to q,B\to s,C\to p$
• C. $A\to q,B\to p,C\to s$
• D. $A\to q,B\to s,C\to s$

Answer 1

The correct option is B $A\to q,B\to s,C\to p$

We know that, $v=\omega \sqrt{A^2-x^2}$
$KE=\frac{1}{2}m{\omega }^2(A^2-x^2)$
$PE=\frac{1}{2}m{\omega }^2\left(x^2\right)$

For $A$:
At $x=\frac{A}{2},$$KE=\frac{1}{2}m{\omega }^2(A^2-{\left(\frac{A}{2}\right)}^2)=\frac{3}{4}\left(\frac{1}{2}m{\omega }^2{A}^2\right)=\frac{3}{4}K{E}_{max}$
$A\to q$

For $B$:
At $x=\frac{A}{2},$$PE=U_{min}+\frac{1}{2}m{\omega }^2{\left(\frac{A}{2}\right)}^2=U_{min}+\frac{1}{4}\left(\frac{1}{2}m{\omega }^2{A}^2\right)=U_{min}+\frac{1}{4}P{E}_{max}$
Since, Data regarding minimum potential energy is not mentioned, we cannot say anything about the Potential energy of the particle.
$B\to s$

For $C$:
Acceleration of particle $a=-{\omega }^{2}x$
At $x=\frac{A}{2};a=\frac{-{\omega }^{2}A}{2}=\frac{a_{max}}{2}$
$C\to p$
Thus, option (b) is the correct answer.