Wave Mechanics Question 531
Question: Two, spring P and Q of force constants $ k _{p} $ and $ kQ( kQ=\frac{k _{p}}{2} ) $ are stretched by applying forces of equal magnitude. If the energy stored in Q is E, then the energy stored in P is
Options:
A) E
B) 2E
C) E/2
D) E/4
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Here, $ k _{Q}=\frac{k _{p}}{2} $ According to Hooke’s law
$ \
Therefore F _{p}=-k _{p}x _{p} $
$ F _{Q}=-k _{Q}x _{Q}\Rightarrow \frac{F _{p}}{F _{Q}}=\frac{k _{p}}{k _{Q}}\frac{x _{p}}{x _{Q}} $
$ F _{p}=F _{Q} $ [Given]
$ \
Therefore \frac{x _{p}}{x _{Q}}=\frac{k _{Q}}{k _{p}} $ ?. (i) Energy stored in a spring is $ U=\frac{1}{2}kx^{2} $
$ \
Therefore \frac{U _{p}}{U _{Q}}=\frac{k _{p}x _{p}^{2}}{k _{Q}x _{Q}^{2}}=\frac{k _{p}}{k _{Q}}\times \frac{k _{Q}^{^{2}}}{k _{^{p}}^{2}}=\frac{1}{2} $
$ [ \
Therefore k _{Q}=\frac{k _{p}}{2} ] $
$ \Rightarrow ,U _{p}=\frac{U _{Q}}{2}=\frac{E}{2} $
$ [\
Therefore U _{Q}=E] $
BETA
