physics
posted by shan .
The motion of a particle is described by x = 10 sin (piet +pie/2 ). At what time ( in second ) is the potential energy equal to the kinetic energy ?
here is what i have done so far where am i going wrong
V=pie10cos(piet + pie/3)
V^2 = pie^2 100cos^2(piet+ pie/3)
KE = mpie5ocos^2 (piet + pie/3)
X^2 =100sin^2(piet +pie/2 )
PE =k50 sin^2(piet +pie/2 )
KE=PE
pie^2 100cos^2(piet+ pie/3) =K50 sin^2(piet +pie/2 )
m/k * pie^2 = tan^2(piet +pie/2 )
take square root of both sides of equation
1/pie * pie = tan(piet + pie/3)
piet + pie/3 =tan1(1)
solving for t i get a negative value where i am i going wrong

You said pi/2 at first, then changed to pi/3
I assume that we are talking about a spring here so
PE = (1/2)kx^2
then
v = 10 pi cos (pi t + pi/2)
v^2 = 100 pi^2 cos^2 (pi t + pi/2)
KE = (1/2) m v^2 = 50 m pi^2 cos^2(pi t +pi/2)
so sort of agree with you
Now PE = (1/2) k x^2 =
(1/2)k 100 sin^2( pi t + pi/2)
so when does PE = KE?
50 m pi^2 cos^2 pi t + pi/2) =50 k sin^2(pi t + pi/2)
m pi^2 cos^2 = k sin^2
tan^2 (pi t + pi/2 ) =pi^2 m/k
tan (pi t +pi/2) = +/ pi sqrt(m/k)
two solutions:
t =1/2 + (1/pi)tan^1(pi sqrt(m/k))
t =1/2  (1/pi)tan^1(pi sqrt(m/k))
Now sqrt (m/k) means something
wo the natural frequency = sqrt (k/m)
= 2 pi fo = 2 pi/T where T is the period of free vibration
so
sqrt (m/k) = T/ 2 pi
so use that
Respond to this Question
Similar Questions

physics
One particle has a mass of 3.00*103 kg and a charge of +7.60 µC. A second particle has a mass of 6.00*103 kg and the same charge. The two particles are initially held in place and then released. The particles fly apart, and … 
PHYSICS helppppp
A 0.2 kilogram baseball is thrown vertically downwards towards to the same horizontal floor. The baseball possesses an initial velocity of 30 meters per second. The height is 10.00m. (A) Calculate the POTENTIAL GRAVITATIONAL ENERGY … 
Physics
The motion of a particle connected to a spring is described by x=10cos(πt). At what time (in s) is the potential energy equal to the kinetic energy? 
Physics
I need help with part b, c, and d please! A 45 g mass is attached to a massless spring and allowed to oscillate around an equilibrium according to: y(t) = 1.4 * sin( 4 * t ) where y is measured in meters and t in seconds.  … 
Physics
I need help with b... I found that k=.750880 for part a but I can't figure out what y would be to get potential energy. A 52 g mass is attached to a massless spring and allowed to oscillate around an equilibrium according to: y(t) … 
Physics
A single conservative force F(x) acts on a 1.6 kg particle that moves along an x axis. The potential energy U(x) associated with F(x) is given by U(x) = 2.9xex/3 where x is in meters. At x = 3.4 m the particle has a kinetic energy … 
Physics
A 245 g particle is released from rest at point A inside a smooth hemispherical bowl of radius 35.0 cm, as shown in Figure 521. (a) Calculate the gravitational potential energy at A relative to B. 1 . (b) Calculate the particle's … 
Physics
The motion of a particle connected to a spring is described by x= 10sin(πt + π/3). At what time (in s) is the potential energy equal to the kinetic energy? 
Physics
The motion of a blockspring system is described by x(t) = A sin(ùt). Find ù, if the potential energy equals the kinetic energy at t = 6.11 s. I'm really not sure how to solve this. It seems that you need more information, either … 
Science (energy)
1. What similarities do nuclear energy and chemical energy share?