Question

In spring-block system match the following columns.

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{(A) If}k\text{(the spring constant) is made 4 times}& \text{(p) Angular speed will become 16 times}\\ \text{(B) If}m\text{(the mass of block) 1s made 4 times}& \text{(q) Potential energy will become 4 times}\\ \text{(C) If}k\text{and}m\text{both are made 4-times}& \text{(r) Kinetic energy will remain unchanged}\\ & \text{(s) None}\end{array}$

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

Answer 1

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

a) We know that, $\omega =\sqrt{\frac{K}{m}}$
If $K$ is made 4 times, then ${\omega }^{\prime }=\sqrt{\frac{4K}{m}}$
$\Rightarrow {\omega }^{\prime }=2\sqrt{\frac{K}{m}}...\left(1\right)$
Potential energy $\left(U\right)=\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}K{x}^2...\left(2\right)$
Using (1),
$U^{\prime }=\frac{1}{2}m{\omega }^{\prime 2}{x}^2$
$=\frac{1}{2}m(4\times \frac{K}{m})x^2$
$=4\left(\frac{1}{2}K{x}^2\right)$
$=4U$
If $K$ is made 4 times, then potential energy will become 4 times.

b) If $m$ is made 4 times. then ${\omega }^{\prime }=\sqrt{\frac{K}{4m}}$
${\omega }^{\prime }=\frac{1}{2}\sqrt{\frac{K}{m}}$
$\therefore$ potential energy $u^{\prime }=\frac{1}{4}\left(\frac{1}{2}m{\omega }^2{x}^2\right)=\frac{U}{4}$

c)If $K$ and $m$ both are made 4 times then $\omega$ remains same $\Rightarrow$ Kinetic energy will remain unchanged.