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A similar calculation for the simple pendulum produces a similar result, namely:

ω max = g L θ max . size 12{ω rSub { size 8{"max"} } = sqrt { { {g} over {L} } } θ rSub { size 8{"max"} } } {}

Determine the maximum speed of an oscillating system: a bumpy road

Suppose that a car is 900 kg and has a suspension system that has a force constant k = 6 . 53 × 10 4 N/m size 12{k=6 "." "53" times "10" rSup { size 8{4} } `"N/m"} {} . The car hits a bump and bounces with an amplitude of 0.100 m. What is its maximum vertical velocity if you assume no damping occurs?

Strategy

We can use the expression for v max size 12{v rSub { size 8{"max"} } } {} given in v max = k m X size 12{v size 8{"max"}= sqrt { { {k} over {m} } } X} {} to determine the maximum vertical velocity. The variables m size 12{m} {} and k size 12{k} {} are given in the problem statement, and the maximum displacement X size 12{X} {} is 0.100 m.

Solution

  1. Identify known.
  2. Substitute known values into v max = k m X size 12{v size 8{"max"}= sqrt { { {k} over {m} } } X} {} :
    v max = 6 . 53 × 10 4 N/m 900 kg (0 . 100 m) . size 12{v size 8{"max"}= sqrt { { {6 "." "53" times "10" rSup { size 8{4} } "N/m"} over {"900"" kg"} } } 0 "." "100"" m"} {}
  3. Calculate to find v max = 0.852 m/s . size 12{v rSub { size 8{"max"} } } {}

Discussion

This answer seems reasonable for a bouncing car. There are other ways to use conservation of energy to find v max size 12{v rSub { size 8{"max"} } } {} . We could use it directly, as was done in the example featured in Hooke’s Law: Stress and Strain Revisited .

The small vertical displacement y size 12{v rSub { size 8{"max"} } } {} of an oscillating simple pendulum, starting from its equilibrium position, is given as

y ( t ) = a sin ωt , size 12{y \( t \) =a"sin"ωt} {}

where a size 12{a} {} is the amplitude, ω size 12{ω} {} is the angular velocity and t size 12{t} {} is the time taken. Substituting ω = T size 12{ω= { {2π} over {T} } } {} , we have

y t = a sin t T . size 12{y left (t right )=a"sin" left ( { {2πt} over {T} } right )} {}

Thus, the displacement of pendulum is a function of time as shown above.

Also the velocity of the pendulum is given by

v ( t ) = 2 T cos t T , size 12{v \( t \) = { {2aπ} over {T} } "cos" left ( { {2πt} over {T} } right )} {}

so the motion of the pendulum is a function of time.

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Why does it hurt more if your hand is snapped with a ruler than with a loose spring, even if the displacement of each system is equal?

The ruler is a stiffer system, which carries greater force for the same amount of displacement. The ruler snaps your hand with greater force, which hurts more.

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You are observing a simple harmonic oscillator. Identify one way you could decrease the maximum velocity of the system.

You could increase the mass of the object that is oscillating.

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Section summary

  • Energy in the simple harmonic oscillator is shared between elastic potential energy and kinetic energy, with the total being constant:
    1 2 mv 2 + 1 2 kx 2 = constant. size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } + { {1} over {2} } ital "kx" rSup { size 8{2} } =" constant"} {}
  • Maximum velocity depends on three factors: it is directly proportional to amplitude, it is greater for stiffer systems, and it is smaller for objects that have larger masses:
    v max = k m X . size 12{v rSub { size 8{"max"} } = sqrt { { {k} over {m} } } X} {}

Conceptual questions

Explain in terms of energy how dissipative forces such as friction reduce the amplitude of a harmonic oscillator. Also explain how a driving mechanism can compensate. (A pendulum clock is such a system.)

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Problems&Exercises

The length of nylon rope from which a mountain climber is suspended has a force constant of 1 . 40 × 10 4 N/m size 12{1 "." "40" times "10" rSup { size 8{4} } "N/m"} {} .

(a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg?

(b) How much would this rope stretch to break the climber’s fall if he free-falls 2.00 m before the rope runs out of slack? Hint: Use conservation of energy.

(c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used.

(a) 1.99 Hz size 12{ "1.99 Hz" } {}

(b) 50.2 cm

(c) 1.41 Hz, 0.710 m

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Engineering Application

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4 . 00 × 10 5 kg size 12{4 "." "00" times "10" rSup { size 8{5} } "kg"} {} on springs that have adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven—the driving force is transferred to the object, which oscillates instead of the entire building. (a) What effective force constant should the springs have to make the object oscillate with a period of 2.00 s? (b) What energy is stored in the springs for a 2.00-m displacement from equilibrium?

(a) 3 . 95 × 10 6 N/m size 12{3 "." "95" times "10" rSup { size 8{6} } "N/m"} {}

(b) 7 . 90 × 10 6 J size 12{7 "." "90" times "10" rSup { size 8{6} } "J"} {}

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Questions & Answers

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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