SATHEE: Simple Harmonic Motion 4 Question 9

Simple Harmonic Motion 4 Question 9

9. 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) $k A$

(d) Zero

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Answer:

Correct Answer: 9. (a)

Solution:

Angular frequency of the system, $ω = \sqrt{\frac{k}{m + m}} = \sqrt{\frac{k}{2 m}}$

Maximum acceleration of the system will be, $ω^{2} A$ or $\frac{k A}{2 m}$ .

This acceleration to the lower block is provided by friction.

Hence, $f_{max} = m a_{max} = m ω^{2} A = m \frac{k A}{2 m} = \frac{k A}{2}$

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Simple Harmonic Motion 4 Question 9

Answer:

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