Spring Practice Problems
Static and oscillating springs KEY
PSYW: Equation, substitute values, solution with label
WATCH YOUR LABELS!!!
Frequency = 1/T
1. A simple pendulum that is 65 cm long is taken to the moon where the gravitational field strength is 1.6 m/s/s compared with that on Earth, which is 10 m/s/s. How is the time period for the swing of the pendulum altered as compared to the period on earth? (Figure out the period on earth and the period on the moon.)
T= ???T = 6.28 √( 0.65 m / 9.8 m/s/s) watch the ( ) !!
L = 0.65 m T = 6.28 √( 0.06632 ) Divide under the √
g = 9.8 m/s/sT = 6.28 (0.25753 sec)take the √
T= 1.61734 sec
T= 1.6 sec2 sig figs
T= ???T = 6.28 √( 0.65 m / 1.6 m/s/s) watch the ( ) !!
L = 0.65 m T = 6.28 √(0.40625) Divide under the √
g = 1.6 m/s/sT = 6.28 (0.6373sec)take the √
T= 4.0027sec
T= 4.0sec2 sig figs
- What is the spring constant for a spring that behaves according to the following graph?
Find the spring constant in Newtons and meters.
Change Displacement to Meters
Choose 2 points, find slope, the slope is the Constant, Label in N/m.
Answer 30 N/m.
3. Based on the diagram below: What is the spring force constant?? (Use the information in the diagram.)
Fsp = Fg =2.0 kg (-9.8m/s/s)19.6 N = - k (.025 m)
Δx = -0.25 m (meters!)Fg = - 19.6 Nk = 78.4 N/m
k= ????
Fg = same as spring force but (-)
m =2.0 kg
ag = -9.8 m/s/s
4. A toy maker requires a spring mechanism to drive an attached component with a period of 0.50s. If the mass of the component is 10g, what must the value of the spring constant, k, be?
T = 0.50 s(0.50 s)2 = 39.44 (0.010 kg) / k
k = ???0.25 s2 = 39.44 (0.010 kg) / ksquare the period
m= 10g = 0.010 kg0.25 s2 = 0.3944 kg / ksimplify numerator
f =(0.25 s2 ) k = 0.3944 kg multiply both sides by “k”
k = 1.5776 N/mdivide by 0.25
k = 1.58 N/mRound to 3 sig figs
5. What is the period of motion for a vertically hung spring that undergoes a series of small oscillations if its force constant is 9.0 N/m and a 2.0 kg mass hangs from its bottom?
T = ???T = 6.28 √ (2.0 kg /9.0 N/m) Be sure to get the ( ) correct!
k = 9.0 N/mT = 2.9604 s
m= 2.0 kgT = 2.9 s (2 sig figs)
6. When a box of unknown mass is placed into the trunk of a car, both rear shocks are compressed a distance of 7.0cm. If we assume the two rear shocks are made from springs, each with a spring constant value of = 35,000N/m, what is the mass of the box? (Assume g = 9.80m/s2).
Fsp = Fsp = - 35000N/m (- 0.07 m) 2,450 N = m (-9.8m/s/s)
Δx = -0.07 m (meters!)Fsp = -2,450 Nm =250 kg
k= 35,000 N/m
Fg = same as spring force but (-)
m =????
ag = -9.8 m/s/s
7. Mass is added to a vertically hanging rubber band and the displacement is measured with the addition of each mass. The recorded data is displayed in the table below. Based on this data, does a rubber band obey Hooke's Law? Explain why or why not?
9. A spring has a spring constant K=100 N/m. When an unknown mass, M, is attached to the spring the mass spring system will oscillate with a frequency of 5 Hz. What is the value of the unknown mass?
T =
k = 100N/m5 Hz = (0.15923) √ ( 100N/m / m)
m= ????25 Hz 2 = (0.025354) ( 100N/m / m) square all factors on both sides
and cancel the square root sign
f =5 Hz(25 Hz 2) m = (0.025354) 100N/mMultiply both sides by “m” from the
equation, NOT the m in the label that means meters!!
(25 Hz 2) m = 25.354 N/mSimplify the right side
m =1.01416 kgDivide through by 25.
m =1. kgRound to 1 sig fig
10. A spring having with a spring constant 1200 N/m is mounted on a horizontal table as shown in the figure. A mass of 3 kg is attached to the free end of the spring pulled sideways to a distance of 2.0 cm and released. Determine the frequency of oscillations
k = 1200N/mf = (0.15923) √ ( 1200N/m / 3kg)
m=3 kgf = (0.15923) √(400 1/s2)simplify under the square root
f =???f = (0.15923) ( 20 1/s)take the square root
f = 3.186 Hzsimplify
f = 3. Hzround