Simple Harmonic Motion of a spring and a mass

In summary, the conversation discusses a spiral spring suspended vertically with a mass attached to its lower end. The spring is extended by 55 mm due to the weight of the mass, and then further extended by an extra 30 mm when the mass is pulled down and released, causing it to oscillate. The conversation also mentions the calculation of the spring constant, time period of oscillations, maximum speed and kinetic energy of the mass, and maximum and minimum tension in the spring. The equations used include Hooke's Law, T=2π/ω, and k.e=1/2mv^2.
  • #1
ibysaiyan
442
0

Homework Statement



A spiral spring of natural length 300mm is suspended vertically from a fixesd point with ita upper end fixed .A mass of 0.150 kg is suspended at rest from the lower end of the spring to increase its length to 355mm.The mass is then pulled down a further distance of 30 mm and released from rest so it osciallates about equlibrium.
Calculate
*spring constant k
*time period of oscillations.
*the max speed and max ke of the mass.
*max and minimum tension in the spring.

Homework Equations


k=fx, time= 2pie/omega,k.e=1/2mv2


The Attempt at a Solution


* for the first bit i got k =0.150x0.085=0.01275Nm/s (not sure)
*how to find time period, as T=2pie/omega or 1/f =( hm..am i missing something here
 
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  • #2


OO i think i got it, i have to use T=2pie[tex]\sqrt{}k/m[/tex]?
if that's the case.. then how do i get max speed for the 3rd bit .. as i don't have amplitude.. and i can't use max speed= A(omega), i am confused as this formula is for max velocity..=/ (or do i simple use v= distance/ time?
 
  • #3


ibysaiyan said:

Homework Equations


k=fx, time= 2pie/omega,k.e=1/2mv2


The Attempt at a Solution


* for the first bit i got k =0.150x0.085=0.01275Nm/s (not sure)
*how to find time period, as T=2pie/omega or 1/f =( hm..am i missing something here

For the first part, look up the Hooke's Law equation again. It's not "k=fx", but is similar to that. You also need to use the weight of the 0.150 kg mass ... and the displacement is not 0.085 m.

For the 2nd question, you're correct that T=2π/ω. If you get ω in terms of k and m, you should be able to work it out then.
 
  • #4


ah thanks for the reply. oo yea. its f= -kx , sorry need to concentrate while typing , so if its not 0.085 displacement.. hm then what is it =/
 
  • #5


Well, the spring was originally 300 mm. Hanging the mass on the spring and letting it come to rest, the new spring length is 355 mm. So the change in length due to the added weight is ____?
 
  • #6


ah sorry been away for sometime ermm.. the change would be 55mm?>but in the question it says.. the spring is further extended by 30 mm ..so that's why i thought total extension is 85mm =/.
 
  • #7


When it is pulled to 85 mm, the force is given by the weight PLUS the force needed to pull it the extra 30 mm. Since we don't know how much that extra force is, we can't plug in for F in F=kx.

When it is just the weight of the mass pulling against the spring, the change in length is 55 mm.
 
  • #8


ah cool.. thanks a lot =]
 

What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion in which an object oscillates back and forth around an equilibrium position, with a constant amplitude and a constant period. This type of motion can be observed in many physical systems, such as a mass attached to a spring or a pendulum.

What is the equation for Simple Harmonic Motion?

The equation for Simple Harmonic Motion is x = A sin(ωt + φ), where x is the displacement of the object from its equilibrium position, A is the amplitude, ω is the angular frequency, and φ is the phase constant. This equation represents the sinusoidal motion of the object as it oscillates back and forth.

What is the relationship between the period and frequency of Simple Harmonic Motion?

The period (T) of Simple Harmonic Motion is the time it takes for one complete oscillation, while the frequency (f) is the number of oscillations per unit time. These two quantities are inversely proportional to each other, meaning that as the frequency increases, the period decreases, and vice versa. This relationship can be expressed as T = 1/f or f = 1/T.

How does the mass and spring constant affect Simple Harmonic Motion?

The mass and spring constant have a direct effect on the period and frequency of Simple Harmonic Motion. A higher mass will result in a longer period and a lower frequency, while a higher spring constant will result in a shorter period and a higher frequency. This is because a heavier mass requires more force to accelerate and a higher spring constant results in a stiffer spring, causing the object to oscillate faster.

What factors can cause Simple Harmonic Motion to deviate from ideal behavior?

Simple Harmonic Motion can deviate from ideal behavior due to factors such as friction, air resistance, and non-ideal spring properties. These external forces can dampen the oscillations and cause the amplitude to decrease over time, resulting in a non-sinusoidal motion. Additionally, non-ideal spring properties, such as stiffness variations, can also affect the period and frequency of the motion.

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