1. A 0.5 kg mass extends a spring 1 cm. What is the frequency of oscillation of
ID: 2208608 • Letter: 1
Question
1. A 0.5 kg mass extends a spring 1 cm. What is the frequency of oscillation of this mass and spring? 2. A fly lands on the mass. This increases the oscillation period by 0.001 second. What is the mass of the fly? 3. A spring has a spring constant of k=600 N/m. a) A 1 kg mass is placed on one of the springs, which is tied to a bracket. How far does it extend? b) What is the force of the spring on the bracket? c) The bracket is replaced with another identical spring. How far does the second spring extend due to the force of the first spring? (This is the case of series springs, as shown in Figure 3). d) What is the total extension of the two springs together? e) The effective spring constant is the total extension divided by the gravitation forceExplanation / Answer
1)Frequency=number of cycle per second or f=1/t
time=2pi*sqr root(mass/spring constant 'k')
so you need to find the spring constant k: k=mg/displacement
k=((0.5kg)*(9.81m/sec^2)) / (.01m)
k= 490.5
now plug into time equ:
t=2pi*sqr root((0.5)/(490.5))=0.2005
now take the inverse of time = 1/t
1/.2005 = 4.98 or roughly about 5 cycles per second
2)1//if it’s just a swinging pendulum then its peropd is T=2pi*(L/g), hence you can see mass of bob does not influence on perpod;
2//if it’s a bouncing on spring then its perdop is T=2pi*(m/k), thence
dT = 0.5*[2pi/(m/k)] *dm/k, hence the mass of fly
dm = dT*k*(m/k) /pi =0.5k*T*dT/pi^2, where dT=0.001s;
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