
Concept explainers
(a)
Prove that the maximum stretching of spring from mean position is x=μsmgk.
(a)

Answer to Problem 60P
It is proved that the maximum stretching of spring from mean position is x=μsmgk.
Explanation of Solution
Write the relation between force on spring and maximum static frictional force.
kx=μsn
Here, k is the spring force, x is the displacement of spring, μs is the coefficient of static friction, and n is the normal reaction force.
Write the expression for n.
n=mg
Here, m is the mass of block and g is the gravitational acceleration.
Rewrite the first equation by substituting the above one.
kx=μsmg
Conclusion:
Rewrite the expression for x.
x=μsmgk
Therefore, it is proved that the maximum stretching of spring from mean position is x=μsmgk.
x=μsmgk.
(b)
Prove that block oscillates about mean point when spring is stretched by x=μsmgk.
(b)

Answer to Problem 60P
It is proved that block oscillates about mean point when spring is stretched by x=μsmgk.
Explanation of Solution
The block will be almost at rest at the break point because of the smaller value of velocity. The block starts to oscillate s soon when the force on block becomes −kx and +μkmg. Write the expression for net force on block while sliding.
Fnet=−kx+μkmg
Here, Fnet is the net force on block,.
Rewrite the above equation by multiplying and dividing the second term with k.
Fnet=−kx+kμkmgk=−k(x−μkmgk)
Express the term μkmgk into a single variable.
x−μkmgk=xrel
Here, xrel is the displacement of block from the point μkmgk.
Rewrite the expression for Fnet by substituting the above relation.
Fnet=−k(xrel)
Conclusion:
From the above calculation, it is evident that block starts to be in simple harmonic motion about mean position on displacing the spring
Therefore, it is proved that block oscillates about mean point when spring is stretched by x=μsmgk.
(c)
Plot the variation of position of block with time.
(c)

Answer to Problem 60P
The plot is
Explanation of Solution
The graph is plotted with time on x-axis and position on y-axis.
The dotted line along x-axis denotes the function xrel=μkmgk. The peak on curve denotes x=μsmgk
Conclusion:
Therefore, the graph is plotted
(d)
Prove that the amplitude of oscillation is A=(μs−μk)mgk.
(d)

Answer to Problem 60P
It is proved that the amplitude of oscillation is A=(μs−μk)mgk.
Explanation of Solution
Write the expression for xrel from part (b).
xrel=x−μkmgk
Rewrite the above expression by substituting μsmgk for x.
xrel=μsmgk−μkmgk=(μs−μk)mgk
Amplitude is the maximum displacement from the mean position only. Rewrite the expression as amplitude.
A=(μs−μk)mgk
Here, A is the amplitude.
On pulling the block to distance x=μsmgk, it will start to oscillate about x=μkmgk.
Conclusion:
Therefore, it is proved that the amplitude of oscillation is A=(μs−μk)mgk.
(e)
Prove that the period of oscillation is T=2(μs−μk)mgvk+π√mk.
(e)

Answer to Problem 60P
It is proved that the period of oscillation is T=2(μs−μk)mgvk+π√mk.
Explanation of Solution
Write the expression for time taken by block to move with the board
T1=2Av
Here, T1 is the time period and v is the speed of motion.
Rewrite the expression by substituting (μs−μk)mgk for A.
T1=2((μs−μk)mgk)v
Time taken by block to move from mean position to an extreme position.
T1/2=12(2π√mk)=π√mk
Here, T1/2 is the time taken to complete the half of SHM (same as that of time needed to spring back the block to original position.
The period of oscillation is the sum of T1 and T1/2.
T=T1+T1/2
Here, T is the time period of oscillation.
Conclusion:
Rewrite the above equation by substituting the relations for T1 and T1/2.
T=2((μs−μk)mgk)v+π√mk=2((μs−μk)mg)kv+π√mk
Therefore, It is proved that the period of oscillation is T=2(μs−μk)mgvk+π√mk.
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Chapter 12 Solutions
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