errors springs

Frequently Made Errors in Mechanics – Springs

[Total: 2    Average: 4/5]

 

1. Springs in Series

“A spring of constant ##k_1## is connected in series with a spring of constant ##k_2##.  What is the spring constant for the composite spring?”

Attempted answer:

When the tension in the first spring is T, the tension in the second spring will also be T.  The two extensions will be ##\frac T{k_1}## and ##\frac T{k_2}##.  Total extension = ##T\left(\frac 1{k_1}+\frac 1{k_2}\right)##.

X Total tension = 2T

Overall spring constant = ##\frac{2T}{T\left(\frac T1{k_1}+\frac 1{k_2}\right)} = \frac{2k_1k_2}{k_1+k_2}##

Tensions in series do not add.  If the tension in a spring (or string) is T then the pull it exerts at each end is T.  What that pull is balanced by at one end, be it a wall or another spring, has no effect on the pull exerted at the other end.
Overall tension = T

Overall spring constant = ##\frac{k_1k_2}{k_1+k_2}##

[Note the analogy with capacitors in series and conductances in series.  Similarly, note the analogy between springs in parallel, capacitors in parallel and conductances in parallel.]

Corollary: If you chop a spring halfway along its length, each half has double the spring constant of the original.

2. Springs as Projectiles

“A spring mass ##m##, constant ##k##, length ##L## stands upright.  It is compressed a distance ##x## and released.  What will be the maximum height ##h## of the mass centre  of the spring?  Assume ##mg \ll kx##.”

Attempted answer 1:

Since ##mg \ll kx##, we can take the spring as being initially at relaxed length.  The energy stored in the compressed spring is ##\frac 12 kx^2##.

X ##\frac 12 kx^2 = mg(h-L/2)##.

Attempted answer 2:

As the spring returns to its relaxed length, it gains gravitational PE.

X ##\frac 12 (kx^2-mgx) = mg(h-L/2)##

In most spring questions, the mass of the spring can be ignored.  Where the spring itself is the projectile, its mass is clearly significant.

In general, the compression in a massive vertical spring will be more towards the bottom of the spring than towards the top, and it is not immediately obvious how uniformly the spring will expand when it is released.  However, since ##mg \ll kx## we ignore those complications.

What we cannot ignore is that at take-off the parts of the spring will not all be moving at the same speed.  If the middle of the spring is rising with speed ##v## at take-off, the part at height ##y## will be rising at speed ##2v\frac yL##.

##\frac 12 kx^2-mgx/2 = \frac 12\int_0^L\frac mL\left(2v\frac yL\right)^2.dy=\frac 23mv^2##

Note this says that of the available energy (lost spring PE minus gained gravitational PE), only three quarters turns into linear KE.  The rest has gone into oscillations within the spring.
##\frac 38( kx^2 -mgx) = mg(h-L/2)##

3. Masses on Vertical Springs

“A mass m falls from height h above a vertical spring of constant k.  What is the maximum compression of the spring?”

Attempted answer:

“PE lost in reaching spring = mgh = KE gained.

At maximum spring compression, the mass will be stationary.

After hitting spring, KE lost = mgh, spring PE gained = ##\frac 12 kx^2##,

X so ##x=\sqrt{\frac{2mgh}k}##”

The catch is that during compression of the vertical spring more gravitational PE is lost.

“Total gravitational PE lost in coming to rest = mg(h+x), spring PE gained = ##\frac 12 kx^2##, ##x=d+\sqrt{d^2+2dh}## where ##d=\frac{mg}k##.”

Note that d here equals the equilibrium compression of the spring with the mass resting on it.

 4. Change in PE

When the extension of a spring changes, the change in PE depends on both the change in extension and the original extension.

“A spring of constant k is at first stretched to an extension ##x_0##.  It is then further stretched by ##x_1##.  What is the increase in its potential energy?”

Attempted answer:

X##\frac 12 k {x_1}^2##”

When extended by ##x_0##, the PE in the spring was ##\frac 12 k {x_0}^2##.  After extending a further ##x_1##, the PE is ##\frac 12 k (x_0+x_1)^2##.

The increase in PE is

##\frac 12 k (x_0+x_1)^2-\frac 12 k {x_1}^2=\frac 12 k (2x_0+x_1)x_1##

Note that we can arrive at the same result by thinking about the average force during the extension:

Work done ##=x_1\left(\frac {kx_0+k(x_0+x_1)}2\right)##

 

Masters in Mathematics. Interests: climate change & renewable energy; travel; cycling, bushwalking; mathematical puzzles and paradoxes, Azed crosswords, bridge

3 replies
  1. QuantumCurt
    QuantumCurt says:

    That was a good write up. I've been tutoring a bit of freshman level physics this last year (during my sophomore year), and I've had to correct some similar errors that people were making. It was made more difficult because these students had not yet completed any E&M, so I was unable to use the analogy of adding resistors in series and parallel. I've found that identifying and helping people with these frequently made errors has really improved my own knowledge of the material.

  2. Greg Bernhardt
    Greg Bernhardt says:

    [SIZE=4][URL=’https://www.physicsforums.com/members/haruspex.334404/’]haruspex[/URL] is on a writing tear at the moment. When I get a chance I’m going to link all his entries together. It’s a wonderful collection.[/SIZE]

  3. Merlin3189
    Merlin3189 says:

    Sec 1 -There seems to be a small typo in the 6[SUP]th[/SUP] line here

    Attempted answer:
    When the tension in the first spring is T, the tension in the second spring will also be T.
    The two extensions will be ##frac{T}{k_1} and frac{T}{k_2}##
    Total extension ## = T(frac{1}{k_1}+frac{1}{k_2})##
    [COLOR=#ff0000]X[/COLOR] Total tension = 2T
    Overall spring constant ## = frac{2T}{T(frac{T}{1}k_1+frac{1}{k_2})}=frac{2k_1k_2}{k_1+k_2}##

    I think it should be
    Overall spring constant ## = frac{2T}{T(frac{1}{k_1}+frac{1}{k_2})}=frac{2k_1k_2}{k_1+k_2}##

    Reference [URL]https://www.physicsforums.com/insights/frequently-made-errors-mechanics-springs/[/URL]

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