A force F = 10 sin nt N acts on a displacement of x = 2 sin (t – n16)m. Determine (a) the work done during the first 6 s; (b) the work done during the first 12s. Solution: Rewriting Eq. (3.7-1) as W=[Fx'dt and substituting F = Fo sin t and x = X sin ( - ) gives the work done per cycle of W=nF0X0 sin ở For the force and displacement given in this problem, Fo = 10 N, X = 2 m, = r/6and the period = 2s. Thus, in the 6 s specified in (a), three complete cycles take place, and the work done is W=3(rF0X0sin p)=37xx10x2 × sin 30°=94.2 N · m The work done in part (b) is determined by integrating the expression for work between the limits 0 and 12s. W=@F0X0[cos 30°S01/2sin nt cos at dt+sin 30°f01/2sin nt dt] =Tx10x2[–0.8664ncos 2nt+0.50(12–sin 2rt47)]01/2 =16.51N m
A force F = 10 sin nt N acts on a displacement of x = 2 sin (t – n16)m. Determine (a) the work done during the first 6 s; (b) the work done during the first 12s. Solution: Rewriting Eq. (3.7-1) as W=[Fx'dt and substituting F = Fo sin t and x = X sin ( - ) gives the work done per cycle of W=nF0X0 sin ở For the force and displacement given in this problem, Fo = 10 N, X = 2 m, = r/6and the period = 2s. Thus, in the 6 s specified in (a), three complete cycles take place, and the work done is W=3(rF0X0sin p)=37xx10x2 × sin 30°=94.2 N · m The work done in part (b) is determined by integrating the expression for work between the limits 0 and 12s. W=@F0X0[cos 30°S01/2sin nt cos at dt+sin 30°f01/2sin nt dt] =Tx10x2[–0.8664ncos 2nt+0.50(12–sin 2rt47)]01/2 =16.51N m
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plz do this problem unable to understand explain all steps related to vibration dolve integration also properly and also find work done during 8 second
![A force F = 10 sin at N acts on a displacement of x = 2 sin (nt – n16)m. Determine (a) the work done during the first 6 s; (b) the
work done during the first 12s.
Solution: Rewriting Eq. (3.7-1) as W=[Fx°dt and substituting F = Fo sin
t and x = X sin (
t- ) gives the work done per cycle of
W=nF0X0 sin ở
For the force and displacement given in this problem, Fo = 10 N, X = 2 m,
= n/6and the period
= 2s. Thus, in the 6 s specified in (a),
three complete cycles take place, and the work done is
W=3(nF0X0sin o)=3rx10x2 × sin 30°=94.2 N •m
The work done in part (b) is determined by integrating the expression for work between the limits 0 and 12s.
W=WF0X0[cos 30°f01/2sin nt cos at di+sin 30°f01/2sin nt dt]
=Tx10x2[-0.8664rcos 2rt+0.50(12–sin 2rt47)]01/2
=16.51N:1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3b1d0714-566e-4db9-bc41-c0271bf1d91b%2Fa4f06299-bb95-4e13-96cb-1486445aa1aa%2Fkkhqu4_processed.png&w=3840&q=75)
Transcribed Image Text:A force F = 10 sin at N acts on a displacement of x = 2 sin (nt – n16)m. Determine (a) the work done during the first 6 s; (b) the
work done during the first 12s.
Solution: Rewriting Eq. (3.7-1) as W=[Fx°dt and substituting F = Fo sin
t and x = X sin (
t- ) gives the work done per cycle of
W=nF0X0 sin ở
For the force and displacement given in this problem, Fo = 10 N, X = 2 m,
= n/6and the period
= 2s. Thus, in the 6 s specified in (a),
three complete cycles take place, and the work done is
W=3(nF0X0sin o)=3rx10x2 × sin 30°=94.2 N •m
The work done in part (b) is determined by integrating the expression for work between the limits 0 and 12s.
W=WF0X0[cos 30°f01/2sin nt cos at di+sin 30°f01/2sin nt dt]
=Tx10x2[-0.8664rcos 2rt+0.50(12–sin 2rt47)]01/2
=16.51N:1
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