Question

In: Physics

Describe static and dynamic methods of determining the force constant of a spring.

Describe static and dynamic methods of determining the force constant of a spring.

Solutions

Expert Solution

Method #1 - Static Approach:

Select a series of masses that give a usable range of displacements. The largest mass should not pull the spring past its elastic limit, for two reasons: (1) beyond the elastic limit there is no well-defined spring constant, and (2) a spring stretched beyond the elastic limit will be damaged.

Clamp the metal rod to the table, and hang the spring from the rod. Decide on a procedure that allows you to measure the distance a spring stretches when an object hangs from it in a consistent manner. Decide how many measurements you will need to make a reliable determination of the spring constant.

Method #2 - Dynamic Approach:

Secure the spring to the metal rod and select a mass that gives a regular oscillation without excessive wobbling. The largest mass you choose should not pull the spring past its elastic limit and the smallest mass should be much greater than the mass of the spring. Practice starting the mass in motion smoothly and consistently.

For both methods, make the measurements that you need to determine the spring constant. DO NOT STRETCH THE SPRINGS PAST THEIR ELASTIC LIMIT (ABOUT 60 CM) OR YOU WILL DAMAGE THEM.


Related Solutions

Briefly describe static linking and dynamic linking. Include the advantages and disadvantages of dynamic linking over static linking.
Briefly describe static linking and dynamic linking. Include the advantages and disadvantages of dynamic linking over static linking.Describe a Windows DLL.Suppose you are writing a DLL and need to export the function void CYBR215(). Show the C syntax to export this function for use by other modules or executables.
In a spring gun system, a spring with a spring force constant 420 N/mN/m  , is compressed...
In a spring gun system, a spring with a spring force constant 420 N/mN/m  , is compressed 0.13 mm . When fired, 80.9 %% of the elastic potential energy stored in the spring is eventually converted into kinetic energy of a 6.10×10−2 kgkg uniform ball that is rolling without slipping at the base of a ramp. The ball continues to roll without slipping up the ramp with 89.6 %% of the kinetic energy at the bottom converted into an increase in...
6. The dynamic spring constant k (in N/m) determined using an oscillating vertical spring is directly...
6. The dynamic spring constant k (in N/m) determined using an oscillating vertical spring is directly related to: - the reciprocal of the slope of a graph of period squared (in s2) as a function of total added mass (in kg) - the slope of a graph of period squared (in s2) as a function of total added mass (in kg)     - the reciprocal of the slope of a graph of period (in s) as a function of total added...
A particle of mass 2.00 kg is attached to a spring with a force constant of...
A particle of mass 2.00 kg is attached to a spring with a force constant of 300 N/m. It is oscillating on a horizontal frictionless surface with an amplitude of 4.00 m. A 7.00 kg object is dropped vertically on top of the 2.00 kg object as it passes through its equilibrium point. The two objects stick together. (a) Does the amplitude of the vibrating system increase or decrease as a result of the collision? decreases increases no change (b)By...
A particle of mass 4.00 kg is attached to a spring with a force constant of...
A particle of mass 4.00 kg is attached to a spring with a force constant of 300 N/m. It is oscillating on a horizontal frictionless surface with an amplitude of 5.00 m. A 9.00 kg object is dropped vertically on top of the 4.00 kg object as it passes through its equilibrium point. The two objects stick together. a) By how much does the amplitude of the vibrating system change as a result of collision? b) By how much does...
A mass m = 3.27 kg is attached to a spring of force constant k =...
A mass m = 3.27 kg is attached to a spring of force constant k = 60.9 N/m and set into oscillation on a horizontal frictionless surface by stretching it an amount A = 0.17 m from its equilibrium position and then releasing it. The figure below shows the oscillating mass and the particle on the associated reference circle at some time after its release. The reference circle has a radius A, and the particle traveling on the reference circle...
A 0.92 kg mass is attached to a light spring with a force constant of 30.9...
A 0.92 kg mass is attached to a light spring with a force constant of 30.9 N/m and set into oscillation on a horizontal friction-less surface. If the spring is stretched 5.0 cm and released from rest, determine the following. (a) maximum speed of the oscillating mass? in m/s ? (b) speed of the oscillating mass when the spring is compressed 1.5 cm? in m/s ? (c) speed of the oscillating mass as it passes the point 1.5 cm from...
A 50.0 g object is attached to a horizontal spring with a force constant of 5.0...
A 50.0 g object is attached to a horizontal spring with a force constant of 5.0 N/m and released from rest with an amplitude of 20.0cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless? (please write out formula used)
A 215 g object is attached to a spring that has a force constant of 72.5...
A 215 g object is attached to a spring that has a force constant of 72.5 N/m. The object is pulled 7.75 cm to the right of equilibrium and released from rest to slide on a horizontal, frictionless table. Calculate the maximum speed of the object. maximum speed: m/s Find the locations of the object when its velocity is one-third of the maximum speed. Treat the equilibrium position as zero, positions to the right as positive, and positions to the...
A frictionless block of mass 1.55 kgkg is attached to an ideal spring with force constant...
A frictionless block of mass 1.55 kgkg is attached to an ideal spring with force constant 350 N/m . At t=0 the spring is neither stretched nor compressed and the block is moving in the negative direction at a speed of 12.9 m/s . a)Find the amplitude. b)Find the phase angle. c)Write an equation for the position as a function of time. a)x=(−x=(−0.858 mm )sin(()sin((15.0 rad/srad/s )t))t) b)x=(−x=(−0.858 mm )cos(()cos((15.0 rad/srad/s )t))t) c)x=(−x=(−15.0m)sin((m)sin((0.858rad/s)t)rad/s)t) d)x=(−x=(−15.0m)cos((m)cos((0.858rad/s)t)rad/s)t)
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT