Answer:
Explanation:
y_1 = (3 mm) sin(x - 3t)
comparing it with standard wave equation
y = A sin( ωt-kx )
we see
ω = -3 , k = -1
velocity = ω / k
= 3
y_2 = (6 mm) sin(2x - t)
we see
ω = -1 , k = -2
velocity = ω / k
= .5
y_3 = (1 mm) sin(4x - t)
we see
ω = -1 , k = -4
velocity = ω / k
= .25
y_4 = (2 mm) sin(x - 2t)
we see
ω = -2 , k = -1
velocity = ω / k
= 2
So greatest velocity to lowest velocity
y_1 = (3 mm) sin(x - 3t) , y_4 = (2 mm) sin(x - 2t) ,y_2 = (6 mm) sin(2x - t) , y_3 = (1 mm) sin(4x - t)
b )
Given the mass per unit length of wire the same , velocity is proportional to
√ T , where T is tension
so in respect of tension in the wire same order will exist for highest to lowest tension .
Otal energy= PEmax on the spring = KEmax of the oject
<span>Total Energy= KE+PE at v or x not maximum. </span>
<span>Using the second line </span>
<span>E= 3/4 E + PE </span>
<span>PE = 1/4 E </span>
<span>PEmax =E = 1/2 kA^2 </span>
<span>PE=E/4 = (1/2)kA^2 / 4= kA^2/8
</span>
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Answer:
Explanation:
Circle motion
Weight=4lb
Radius=3ft
Velocity Vb= 6ft/s
Velocity Vr=2ft/s
Velocity Vo=12
V2=√Vo²+Vr²
12=√Vo²+2²
Square both side
144=Vo²+4
Vo²=140
Vo=11.83ft/s
Applying conservation of angular momentum
Ha1=Hb2
MbVbr1=MbVor2
r2=Vbr1/Vo
r2=6×3/11.83
r2=1.52ft
The require time is written as.
∆r=Vrt
t=∆r/Vr
t=r1-r2/Vr
t=3-1.52/2
t=0.74sec
