Spring Problems with Oscillations
Homework worksheet:
KEY
1. What is the period of motion for a vertically hung spring that undergoes a series of small oscillations if it's force constant is 9.0 N/m and a 2.0 kg mass hangs from its bottom?
T = ???T = 6.28 √ (2.0 kg /90 N/m) Be sure to get the ( ) correct!
k = 90 N/mT = 0.9361 s
m= 2.0 kgT = 0.94 s (2 sig figs)
f =
2. What is the period of motion for a vertically hung spring near the Earth's surface that undergoes a series of small oscillations if its force constant is 64 N/m and a 39.2 Newton mass hangs from its bottom
Note: Newton is the unit for Force of gravity not mass!!
Fg = 39.5 N39.65 N = m (-9.8 m/s/s)
m =??m= 4.02 kg
ag = -9.8 m/s/s
T = ??/T = 6.28 √ (4.02 kg /64 N/m) Be sure to get the ( ) correct!
k = 64 N/mT = 1.5739 s
m= ??? = 4.02kgT = 1.6 s (2 sig figs)
f =
- 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 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
4. 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
5. A spring of constant k is hung vertically and not extended. A mass 560 grams is attached to the spring and it stretches a distance of 56 cm. (a) Find k. (b) Find the frequency of the motion when the spring is released.
Fsp = Fg =0.500kg (-9.8m/s/s)+4.9N = - k (-0.56 m )
Δx = -0.56m (meters!)Fg = -4.9Nk = 8.75 N/m
k= ???N/m
Fg =
m =0.500 kg
ag = -9.8 m/s/s
T =
k = ?????=8.75 N/m
m=
f =?????f = (1/6.28) √ (90 N/m /2.0 kg) Be sure to get the ( ) correct!
f = (1/6.28) √(45 s2) Simplify under the √
f = (1/6.28) 6.708 1/stake the square root
f = 1.068 HzMultiply and get the answer
6. A spring with spring constant k = 10.0 N/m is attached to an object of mass m = 0.300 kg. One third of the spring is cut off. What is the frequency of the oscillations when the “new” spring-mass is set into motion?
The spring constant would stay the same—the stiffness of the string is what the spring constant tells us.
7. A 0.200 kg mass is launched vertically upward by a compressed spring. The mass is pushed down onto the originally relaxed spring compressing it 0.10 m, diagram A. It is then released from rest and the spring launches it upward, diagram B. The spring constant is 500. N/m. What is the period of the oscillation?
T = ???T = 6.28 √ (0.2 kg /500 N/m) Be sure to get the ( ) correct!
k = 500 N/mT = 6.28 √ (100 sec squared)Simplify under the √
m= 0.200 kg.T = 6.28 (10 sec)take the square root
f =T = 62.8 sec 3 sig figs,