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hc Verma solutions
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Asked by mlakshmi22872 | 01 Mar, 2021, 07:24: PM
answered-by-expert Expert Answer
Part (a)
 
F = k x
 
where F = 10 N is applied force , k = 100 N/m  is spring constant and x is compressed distance .
 
Compressed distance x = F / k = ( 10 / 100 ) = 0.1 m
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Part (b)
 
Potental energy U = 0
 
Kinetic energy K = (1/2) m v2 = 0.5 × 1 × 2 × 2 = 2 J
 
where m is mass and v is speed of mass at the instant it was struck
 
Sum of potential energy and kinetic energy , U + K = 2 J
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Part (c)
 
Angular speed ω is given as
 
begin mathsize 14px style omega space equals fraction numerator 2 pi over denominator T end fraction space equals space square root of k over m end root end style
where T is  period of oscillation
 
begin mathsize 14px style T space equals space 2 pi square root of m over k end root space space equals space 2 pi space square root of 1 over 100 end root space equals space fraction numerator 2 pi over denominator 10 end fraction space equals space 0.63 space s end style
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Part (d)
 
Amplitude a is determined from maximum potential energy that equals maximum kinetic energy .
 
Maximum kinetic energy is 2J as calculated from part (b)
 
(1/2) k a 2 = 2 J
 
a = { 4 / k }1/2 = { 4/ 100 }1/2 = 0.2 m
 
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Part (e)
 
Potential energy at left extreme is maximum potential energy Umax
 
Umax = (1/2) k a2 = (1/2) × 100 × 0.2 × 0.2  = 2 J
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Part (f)
Potential energy at right extreme is maximum potential energy Umax
 
Umax = (1/2) k a2 = (1/2) × 100 × 0.2 × 0.2  = 2 J
------------------------------------------------------------------
 
Answer to Part (b) is total energy of oscillating spring-mass system.
 
Answer to Part (e) and (f) are maximum potential energy at maximum displacement
 
In Part (b) potential energy is zero . In part (e) and part (f) kinetic energy is zero .
All of them are maximum energy of the system and they are same not different.
 
There is no violation of conservation of energy
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