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If a spring mass system is placed on a horizontal smooth surface. Then find the net work done by spring force after one time period(assume suitable variables) 

Asked by Pjdrd24680 15th September 2018, 8:06 AM
Answered by Expert
Answer:
Let us assume the mass m is subjected to SHM due to spring force. 
 
dispalcement x at any time t is given by,   x = a sin(ωt ) .......................(1)
 
where a is displacement, ω is angular speed.
 
speed at time t is given by, begin mathsize 12px style fraction numerator d x over denominator d t end fraction space equals space a omega space cos open parentheses omega t close parentheses space......................... left parenthesis 2 right parenthesis end style
acceleration at time t is given by, begin mathsize 12px style fraction numerator d squared x over denominator d t squared end fraction space equals space minus space a omega squared sin open parentheses omega t close parentheses space......................... left parenthesis 3 right parenthesis end style
force F = mass× acceleration begin mathsize 12px style equals space m cross times fraction numerator d squared x over denominator d t squared end fraction space equals space minus space a cross times omega squared cross times m cross times sin open parentheses omega t close parentheses space......................... left parenthesis 4 right parenthesis end style
workdone dW for small displacement dx is given by,
 
 dw = F×dx begin mathsize 12px style equals space minus space a cross times omega squared cross times m cross times sin open parentheses omega t close parentheses space cross times d x space equals space minus space a cross times omega squared cross times m cross times sin open parentheses omega t close parentheses space cross times a cross times omega cross times cos open parentheses omega t close parentheses cross times d t space......................... left parenthesis 5 right parenthesis end style
eqn.(5) is rewriteen as, begin mathsize 12px style F cross times d x space equals space fraction numerator negative a squared cross times omega cubed cross times m over denominator 2 end fraction sin open parentheses 2 omega t close parentheses d t space equals space fraction numerator negative a squared cross times omega squared cross times m over denominator 4 end fraction sin open parentheses 2 omega t close parentheses cross times left parenthesis 2 omega d t right parenthesis end style ........................(6)
Let us substitutute θ = 2×ω×t in (6) and rewrite as, begin mathsize 12px style F cross times d x space equals space space fraction numerator negative a squared cross times omega squared cross times m over denominator 4 end fraction sin open parentheses theta close parentheses cross times d theta space........................ left parenthesis 7 right parenthesis end style
workdone for one time period,  begin mathsize 12px style contour integral F cross times d x space equals space space fraction numerator negative a squared cross times omega squared cross times m over denominator 4 end fraction integral subscript 0 superscript 4 pi end superscript sin open parentheses theta close parentheses cross times d theta space........................ left parenthesis 8 right parenthesis end style
Note the limit of integration :- when t = T,   θ = 2×ω×t = 2×ω×T = 4π , where T is one time period
 
when a sine function is integrated from 0 to 4π, i.e. two full cycles, net result is zero.
 
hence net workdione for a time period is zero.
Answered by Expert 17th September 2018, 12:27 PM
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