Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in figure. When the mass is displaced from equilibrium position by a distance x towards right, find the restoring force.

There are two identical springs, A and B, that have the same spring constant k. Spring A is stretched twice as far as spring B. The magnitude of the force due to spring A is two times larger than that due to spring B.

A net force with magnitude 5 N is applied to object A. During the subsequent motion, the average speed is 1 m/s. Object B is identical to object A. If object B has the same force applied to it, but the power is twice as large, the average speed of object B

Feb 28, 2018 Two identical springs each of force constant k are connected in series and parallel so that they supports a mass M. find the ratio of the time period of the mass in the two system

Pre-class Exam One. There are two identical springs, A and B, that have the same spring constant k. Spring A is stretched twice as far as spring B. The magnitude of the force due to spring A is two times larger than that due to spring B.

In the figure, two identical springs have unstretched lengths of 0.25 m and spring constants of 300 N/m. The springs are attached to a small cube and stretched to a length L of 0.36 m as in Figure A. An external force P pulls the cube a distance D = 0.020 m to the right and holds it there.

Oct 02, 2013 The ends of two identical springs are connected. Their unstretched lengths l are negligibly small and each has spring constant k. After being connected, each spring is stretched an amount L and their free ends are anchored at y = 0 and x = ±L as shown (Figure 1) .

The applied force will produce the same tension, Fap in both springs. Hence each spring will increase in length by Δx = Fap/k ,or the entire system will undergo a length change Δxtotal = Δx = Fap/k = Fap/keq, and therefore 1/keq = 1/2k, or keq = 2k.

Oct 02, 2013 The ends of two identical springs are connected. Their unstretched lengths l are negligibly small and each has spring constant k. After being connected, each spring is stretched an amount L and their free ends are anchored at y = 0 and x = ±L as shown (Figure 1) .

Dec 15, 2016 A constant force vecF is exerted on the rod so that remains perpendicular to the direction of the force. So that the springs are extended by the same amount. Alternatively, the direction of force could be reversed so that the springs are compressed. This system of two parallel springs is equivalent to a single Hookean spring, of spring constant k.

Two identical springs, each with spring constant k, are attached in parallel to a mass, which is then set into simple harmonic motion. What would be the spring constant of a single spring which would result in the same frequency of oscillation as the parallel springs? a. k b. 2k c. k/2 d. k

Nov 17, 2007 1.When you pull with a constant force (F) what happens to delta x if the single spring is replaced with a) two identical springs in series? b) two identical springs in parallel? Assume all springs have the same spring constant and always compare to the single spring case? 2. The only equations i

Apr 22, 2013 Two springs have the same spring constant k = 130N/m but different equilibrium lengths, 3.0cm and 5.0cm . They are arranged as shown in the figure, then a block is pushed against them, compressing both to a length of 2.5cm .(click the link for the pic of the problem)

Aug 21, 2014 When you apply a force to springs in series, the force is equal in each spring, so each spring compresses (say) by the same amount. So the total system compresses by 3x what one spring would compress by. Since k is a stiffness, the resultant sys

Jan 09, 2008 Now a spring has a force given by: F = kx so the work done to stretch a spring is W = 1/2kx^2. If the springs are stretched the same distance, then clealry the stiffer spring has more work done on it because k will be greater for teh stiffer spring. If force is constant, the stiffer spring is stretched less ( x = F/k).

Two springs a and b having spring constants Ka and KB with Ka=2kb are stretched by applying a force of equal magnitude .If energy stored in spring A is Ea what is the energy stored in spring B

Two identical masses are hung over two pulleys of identical mass M and radius R, but different mass distributions. The mass hung from pulley 1 falls slower than the mass hung from pulley 2. For which pulley is the net force on the falling mass larger?

A transverse wave with a speed of 50 m/s is to be produced on a stretched spring. If the string has a length of 5.0 m and a mass of 0.060 kg, what tension on the string is required. When two waves overlap in space the displacement of the wave is the sum of the

COUPLED OSCILLATORS half-spring is twice that of a full spring (because a half-spring is twice as sti as the corresponding full spring, since it stretches only half as much for a given applied force). That is why the frequency in this case is ! 2 = p (k+ 2k0)=m. This motion is the second normal mode of oscillation.

An object with a mass of 20.1 kg is hung from two identical springs vertical. The springs both stretch 0.23 meters from equilibrium with the attached mass. One of the those springs is placed on a horizontal surface without friction and the mass of 20.1 kg . asked by Chelsea on November 15, 2012; Physics

Apr 17, 2013 If two waves traveling in the same direction overlap and thus interfere, which best describes the interference if the waves have a phase difference of 2.0 wavelengths? Use the 30 degree angle and the 5.0cm to find the height of the triangles. b. Knowing that the force on the 1.0nC charge does not have a h. 3 answers (see the figure

Springs--Two Springs in Series The force is the same on each of the two springs. Therefore (1) Solving for in terms of , (2) We are looking for the effective spring constant so that (3) where is the total displacement of the mass. Equating (3) with the right side of (1) and substituting into (2) gives

In response to a request for proposals*, one company states their guard rails are perfect for the job. Each section of their guard rails consists of two springs, each having a force constant 3.13400 105 N/m with a maximimum distance of compression of 0.614 m.

Justification: We learned from question 5 that two springs in parallel will stretch half as much as a single spring with the same mass attached. It requires twice as much force to stretch the two springs. The two springs are “stronger” and have twice the spring constant. Solution () 2 2 22 T T Ts F k

Obviously, the effective force constant must be related . asked by aadsa on July 31, 2013; physics. A 7000-kg freight car rolls along rails with negligible friction. The car is brought to rest by a combination of two coiled springs as illustrated in the figure below. Both springs are described by Hooke's law and have spring . asked by Anonymous on March 23, 2012; Physics helpp please! A 8000-kg freight car rolls

Answer to 2. Three identical springs each have the same spring constant k. If these three springs are attached end to end forming

Mar 23, 2012 I try to find the movement of classic helicoids and the sum of energy with 3 different pressures. For me when two identical helicoids turn in the same direction around their axis, they don't move up/down. In fact, when I see helicoids turning I see

Mar 27, 2016 Answer: The spring constant determines how far the spring will stretch for a given applied force: F=kx→k=F x. If we place the same mass on the two springs, which means we have placed the same force on them, the one that stretches least has the largest spring constant.

I'm assuming that when you say series, the springs are attached to each other, forming a long chain. First off, for this type of question, you need to remember the fact that the force from a spring comes from how long it is stretched multiplied by

Feb 04, 2015 Two springs with the same unstretched length, but different force constants k1 and k2 are attached to a block with mass m on a level, frictionless surface. Calculate the effective force constant keff in each of the three cases depicted in the figure (Figure 1) .

Jun 22, 2011 Now, three masses m1 = 4.6 kg, m2 = 13.8 kg and m3 = 9.2 kg hang from three identical springs in a motionless elevator. The springs all have the same spring constant that you just calculated above. What is the force the top spring exerts on the top mass? I've already calculated that spring constant is 300N/m. (And I know this is correct via a checker.)

Given that the force constants of the two springs are and , find the period of oscillation of the system. Answer: Let and represent the extensions of the first and second springs, respectively. Let and be the magnitudes of the forces exerted by the first and second springs, respectively.

The two forces have the same magnitude. Since the force the spring exerts on you is equal in magnitude to your weight, you exert a force equal to your weight on the spring, compressing it. The change in length of the spring is proportional to your weight. Spring scales use a spring of known spring constant and provide a calibrated readout of

TA = 2TB and the systems’ springs have identical force constants, it follows that the systems’ masses are related by (a) mA = 4mB, (b) mA =mB 2 , (c) mA = mB/2, (d) mA = mB/4. Picture the Problem We can use T =2π m k to express the periods of the two mass-spring systems in terms of their force constants.

Two identical springs with spring constant 50 N/m support a 5.0 N weight as in the picture below. What is the change in length of each spring when the weight is hung on the springs? A 2.0-kg block situated on a frictionless incline is connected to a light spring (k = 100 N/M), as shown.

Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown below Show that the mass executes simple harmonic motion when displaced from its rest position on either side Also, find the period of oscillations - Physics -

Example B. (Identical Spring Rates) You have two identical springs with a spring rate of 30 lbf/in (pounds of force per inch). To calculate the rate these two provide together, simply multiply the rate by 1/2 (0.5). The formula for this calculation is also provided below. The resulting Keq is the new rate for the two springs in series.

C. Springs - Two Springs in Series Consider two springs placed in series with a mass m on the bottom of the second. The force is the same on each of the two springs. Therefore F = −k 1 x 1 = −k 2 x 2 (C-1) Solving for x 1 in terms of x 2, we have: 2 1 2 1 x k k x = (C-2) The force exerted on the mass can also be written as: F = −k eff ()x 1 + x 2 k

question_answer2) A mass m is suspended by means of two coiled spring which have the same length in unstretched condition as in figure. Their force constant are k1 and k2 respectively. Their force constant are k1 and k2 respectively.

the spring constant by: slope = 4⇡2 k (9.5) So the spring constant can be determined by measuring the period of oscillation for di↵erent hanging masses. This is the second way that k will be determined today. 9.5 In today’s lab Today you will measure the spring constant ( k)ofagivenspringintwo ways.

Question: Two identical springs with spring constant 50 N/m support a 5.0 N weight as in the picture below. A 2.0-kg block situated on a frictionless incline is connected to a light spring (k = 100 N/M), as shown. The block is realeased from rest when the spring is

1. A particle of mass 1.8 kg is attached between two identical springs on a horizontal, frictionless tabletop. Both springs have spring constant k and are initially unstressed. (a) The particle is pulled a distance x along a direction perpendicular to the initial configuration of the springs as shown in Figure 1. Show that the force exerted by the

12. Three identical springs each have the same spring constant k. If these three springs are attached end to end forming a spring three times the length of one of the original springs, what will be the spring constant of the combination? a. k b. 3 k c. k /3 d. 1.73 k 13.2 Elastic Potential Energy 13. A 0.20-kg object is oscillating on a spring with a spring constant of k = 15 N/m.

Mar 14, 2009 The ends of two identical springs are connected. Their unstretched lengths l are negligibly small and each has spring constant k. After being connected, both springs are stretched an amount L and their free ends are anchored at y=0 and x= (plus minus)L as shown (Intro 1 figure) . The point where the springs are connected to each other is now pulled to the position (x,y).

Variables that have been kept constant (for all parts) include the spring that was used and the person who flicked the spring. We tried to keep ALL variables constant for all four measurements in each part. The dependent variable is time. Risk Assessment: The risk is the same as Investigation 1, as it too involves stretched springs.

Force constants have been determined from the skeletal vibration frequencies of trisilylamine, N(SiH 3) 3, and disiloxane, O(SiH 3) 2.If the stretching interaction force constant involving two Si—X bonds is the same in the amine and in the ether, then the SiOSi angle is ⩽ 150°.

Nov 01, 2012 Two particles, each of mass M, are hung between three identical springs. Each spring is massless and has spring constant k. Neglect gravity. The masses are connected as shown to a dashpot of negligible mass. The dashpot exerts a force of bv, where v is the relative velocity of its two ends. The force opposes the motion.

Two identical objects are pressed against two different springs so that each spring stores 55.0J of potential energy. The objects are then released from rest. One spring is quite stiff (hard to compress), while the other one is quite flexible (easy to compress).

The spring constant of the half spring should double. Consider a force F. This force is applied to a spring of length L and it causes the spring to extend by X distance. Using this we get, K the spring constant. Where K= F/X. If the spring is cut

force is the analog of voltage. velocity is the analog of current. Keeping this in mind, consider two springs connected in mechanical parallel and note that the velocity (rate of change of displacement) for each spring is identical. But recall, in this analogy, velocity is the analog of current.

The applied force will produce the same tension in both springs. Hence spring one will increase in length by δ x1 = Fap / k1, spring two will increase in length by δ x2 = Fap / k2, and the entire system will undergo a length change δ xtotal = δ x1 + δ x2 = Fap / k1 + Fap / k2 = Fap / keq,

The aim of this investigation is to examine the effect on the spring. constant placing 2 identical springs in parallel and series. combination has and how the resultant spring constants of the parallel. and series spring sets compare to that of a lone spring with identical. spring constant.

Therefore each spring extends the same amount as an individual spring would do. The combination therefore is more 'stretchy' and the effective spring constant for the combination will be half that of a single spring for two in series, a third for three in series etc. Springs in parallel. The weight is supported by the combination.

10 An oscillating block –spring system takes 0.75 s to begin repeating its motion. Find (a) the period, (b) the frequency in hertz, and (c) the angular frequency in radians per second. 11 In Fig. 15 -29, two identical springs of spring constant 7580 N/m are attached to a block of mass 0.245 kg.

Two springs have force constants k1 and k2(k1>k2).On which spring is more work done,if(i)they are stretched by the same force (ii) they are stretched by the same amount?

Physics 207: Lecture 17, Pg 18 Lecture 17, Statics Exercises 4 and 5 1. A hollow cylindrical rod and a solid cylindrical rod are made of the same material. The two rods have the same length and outer radius. If the same compressional force is applied to each rod, which

Where K= F/X. If the spring is cut into half and the same force F is applied to the half spring, the distance it should stretch is X/2, half of the original deformation. From this the new spring constant, K2 becomes K2= F/(X/2) or 2* (F/X) which is double of K. To understand why this happens, we have to consider the way force acts on a spring.

Now consider three springs set up in series as shown. (Figure 2) The spring constants are k1, k2, and k3, and the force acting to the right again has magnitude F. Part B Find the spring constant k? of the three-spring system. Express your answer in terms of k1, k2, and k3.

Two identical springs are attached to two different masses, MA and MB, where MA is greater than MB. The masses lie on a frictionless surface. Both springs are compressed the same

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here! Two springs with the same unstretched length, but different force constants and are attached to a block with mass on a level, frictionless surface. Calculate the effective force constant in each of the three cases depicted in the figure.

displacement ( ∆x) is called the spring constant (k) and can be written as follows: k = ∆F ∆x (2) Today’s experiment will test this relationship for a large spring. By hanging different masses from the spring we can control the amount of force acting on it. We can then measure for each applied weight, the amount that the spring "stretches”.

Two identical objects are attached to identical springs, hence they both have mass m attached to a spring with spring constant k, so the periods are the same and are equal to T 1 =T2 = 2π m k. If one of the springs has a bigger mass attached to it, the period will be longer; if m2 > m1 then T2 > T1 Changing the spring constant also changes the period, if k2 > k1 then T2 < T1 5

We ran the experiment two ways, static and dynamic, by with F being the force, k being the spring constant, and x being the displacement. We tested to see if the cord behaved in accordance to Hooke’s Law for springs, which was the third and final purpose of the type spring would have the same k-value no matter how much of the spring

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