Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 1 (Q.No. 20)
20.
The maximum velocity of a body vibrating with a simple harmonic motion of amplitude 150 mm and frequency 2 vibrations/sec, is
Discussion:
8 comments Page 1 of 1.
Chuck said:
6 years ago
Displacement = Amplitude x sin(angular frequency x time).
y = Asin(ωt) where y = displacement, A = amplitude,
ω = angular frequency, t = time.
Velocity v = ωAsin(ωt).
and,
v(max)= Aω,
Where ω = 2πf,
f= frequency.
ω = 2*3.14*2 = 12.56.
Therefore, v = 0.15*12.56 = 1.884.
y = Asin(ωt) where y = displacement, A = amplitude,
ω = angular frequency, t = time.
Velocity v = ωAsin(ωt).
and,
v(max)= Aω,
Where ω = 2πf,
f= frequency.
ω = 2*3.14*2 = 12.56.
Therefore, v = 0.15*12.56 = 1.884.
Vidhya said:
7 years ago
Nice explanation. Thank you @Abhay.
(1)
Abraham sarker said:
7 years ago
Thank you @Abhay.
Prassu said:
7 years ago
Thank you @Abhay.
Subha said:
8 years ago
Thank you @Abhay.
Abhay said:
10 years ago
Displacement = Amplitude x sin(angular frequency x time).
y = Asin(ωt) where y = displacement, A = amplitude,
ω = angular frequency, t = time.
Velocity v = ωAsin(ωt).
and,
v(max)= Aω,
Where ω = 2πf,
f= frequency.
ω = 2*3.14*2 = 12.56.
Therefore, v = 0.15*12.56 = 1.884.
y = Asin(ωt) where y = displacement, A = amplitude,
ω = angular frequency, t = time.
Velocity v = ωAsin(ωt).
and,
v(max)= Aω,
Where ω = 2πf,
f= frequency.
ω = 2*3.14*2 = 12.56.
Therefore, v = 0.15*12.56 = 1.884.
Dona said:
1 decade ago
Max velocity of SHM is given by;
Vmax = rw.
= w(in radians)*0.15.
= 12.566*0.15 = 1.8849 m/s.
Vmax = rw.
= w(in radians)*0.15.
= 12.566*0.15 = 1.8849 m/s.
Nitesh sankpal said:
1 decade ago
v = Aw sin(wt) where A = 150mm and w = 2*3.14*f = 2*3.14*2 = 12.56.
Now, v=150*12.56 = 1884mm/sec = 1.884m/sec.
Now, v=150*12.56 = 1884mm/sec = 1.884m/sec.
(1)
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