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## [STPM] Physics Daniel Member   Join Date: Jun 2004 Posts: 183 #101 09-09-2004 A spring with a load moving up and down. Why we say that the system are performing SHM? And why we need to relate it with w, that is the angular velocity? The spring is just moving up and down and does not perform any circular motion right? But why we need to relate it with angular velocity? Can someone explain it further?  yekban81 Member   Join Date: Mar 2004 Posts: 296 #102 10-09-2004 SHM-Single Harmonic System..I forget liao the definition. Is there any other type of harmonic system that you learn in your STPM Physics? The motion of spring with load is characterised by using the displacement and time. Let us assume that this motion is a perfect harmonic motion; no damp effect, no friction. Hence this spring moves up and down at a similar amplitude all the time in a certain frequency. When you plot the displacement and time in a single graph, you get a curve which is very similar to the sin/cos curve. In fact it is in the form of equation with sin/cos function. w is the value of theta per time in radian per second, thus it is named as "angular velocity". From the equation: S=Asin(wt) where wt=theta To me, it is just a constant value in a sin/cos equation which represent the motion of a particular spring with a particular load. w is a parameter created to express theta in terms of time, t.  cooldownguy86
Super Junior Member
#103 18-09-2004 i have a question guys. hope you can help.

Quote:
 State Newton's second law of motion. (no problem) When an object moves through a fluid it experiences a retarding force due to turbulence. For a sphere of radius r moving with a speed v in a fluid of density p, the retarding force is given by F = kpr^2v^2, where k is a constant. By relating the retarding force to the transfer of momentum between the sphere and the fluid, explain why the force F is directly propotional to pr^2v^2
it's hard...i can't understand it. thanks in advance  doodle Super Junior Member   Join Date: Feb 2005 Posts: 4 #104 05-02-2005 Let me try... The movement of the sphere in fluid displaces the fluid. Force is defined as the rate of change of momentum, F = d(mv)/dt. Since velocity is constant, then F = v.dm/dt. The mass of fluid displaced when the sphere moves x distance is m = p.V = p.pi.r^2.x. Thus F = v.p.dV/dt = v.p.pi.r^2.dx/dt = pi.p.r^2.v^2 since v = dx/dt.  bush
ReCom Staff
Wiki Contributor
#105 13-04-2005 Quote:
 Originally Posted by Daniel A spring with a load moving up and down. Why we say that the system are performing SHM? And why we need to relate it with w, that is the angular velocity? The spring is just moving up and down and does not perform any circular motion right? But why we need to relate it with angular velocity? Can someone explain it further?
SHM is an occillation or motion such that the acceleration of the system is directly proportional and oppositely directed to its displacement.( acceleration is always directed to its equilabrium point)

for omega w of a SHM, it is called angular frequency, and not angular velocity though they have the same units and dimensions, angular velocity is used in circular motion separately. its just terms......terms to screw up the life of a student  Mark Super Junior Member   Join Date: Jun 2005 Posts: 2 #106 07-06-2005 please help me to solve this question A car is travelling at 13 m/s towards some traffic lights..when the driver sees the lights change to red,the car distance is at a distance 25m from the stop line. The reaction time of the driver is 0.7s and the condition of the road does not permit the car to decelerate greater than 4.5m/s^2. With the brakes fully applied,how far from the stop line is the car when it stops? On which side of the line is the car? Answer:2.9m after the stop line  littlebigone Slightly Senior Member   Join Date: Apr 2003 Posts: 867 #107 09-06-2005 i used a velocity-time graph to solve this question. velocity times time gives you distance, therefore area under the curve that plots velocity will give you the distance travelled. in this problem, i assumed constant deceleration, therefore time taken to get from 13 to 0 is 2.888 ~ 2.9 secs (13/.5) Since we assume constant deceleration, we can easily imagine that a straight line graph with negative slope, with y axis plotting velocity and x axis plotting time. the line cuts the axis at (0,13) and (2.9,0), so we have right angle triangle with sides 13 and 2.9. The area of this triangle is 18.85 = 0.5 x 13 x 2.9. But the driver takes 0.7 secs to react and so travels 9.1 m before starting to brake. The total distance travelled is 27.95 m so the car stops 2.95 meters after the stop line. I think the errors are the result of rounding. Let me know if this solution works for you. __________________ Life is what happens while you're busy making other plans ~John Lennon  Mark Super Junior Member   Join Date: Jun 2005 Posts: 2 #108 09-06-2005 good thx littlebigone...i think you have done it correctly..  yangyang Super Junior Member   Join Date: Jun 2005 Posts: 2 #109 12-06-2005 hey guys, speaking of physics, i do hav a question dat i dunno (actually, according 2 my fren it's quite ez)....pls help me, thx 1.) A man runs at a speed of 4.0m/s to overtake a bus. When he is 6.0m behind the door (at t=0), the bus moves forward and continues with a constant acceleration of 1.2m/s2 a.) How long does it take for the man to gain the door? b.) If in the beginning he is 10.0m from the door, will he (running at the same speed) ever catch up? If possible, pls show da working 4 me...thx u...  lyt87
Less Junior Member
#110 17-06-2005 Quote:
 Originally Posted by yangyang hey guys, speaking of physics, i do hav a question dat i dunno (actually, according 2 my fren it's quite ez)....pls help me, thx 1.) A man runs at a speed of 4.0m/s to overtake a bus. When he is 6.0m behind the door (at t=0), the bus moves forward and continues with a constant acceleration of 1.2m/s2 a.) How long does it take for the man to gain the door? b.) If in the beginning he is 10.0m from the door, will he (running at the same speed) ever catch up? If possible, pls show da working 4 me...thx u...
1a) v=4m/s s=6
s=vt, 4t=6
t = 1.5

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