Simple Harmonic Motion Ques 48
- A block $P$ of mass $m$ is placed on a horizontal frictionless plane. A second block of same mass $m$ is placed on it and is connected to a spring of spring constant $k$, the two blocks are pulled by a distance $A$. Block $Q$ oscillates without slipping. What is the maximum value of frictional force between the two blocks?
(2004, 2M)
(a) $k A / 2$
(b) $k A$
(c) $\mu _s m g$
(d) Zero
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Answer:
Correct Answer: 48.(a)
Solution:
Formula:
- Angular frequency of the system, $\omega=\sqrt{\frac{k}{m+m}}=\sqrt{\frac{k}{2 m}}$
Maximum acceleration of the system will be, $\omega^{2} A$ or $\frac{k A}{2 m}$.
This acceleration to the lower block is provided by friction.
Hence, $f _{\max }=m a _{\text {max }}=m \omega^{2} A=m \frac{k A}{2 m}=\frac{k A}{2}$
BETA
