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hc Verma solutions
Asked by mlakshmi22872 | 01 Mar, 2021, 07:24: PM
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

where T is  period of oscillation

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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
----------------------------------------------------------

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
Answered by Thiyagarajan K | 02 Mar, 2021, 10:19: AM

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