A block rests on a horizontal frictionless table. It is attached to a spring and
ID: 1418819 • Letter: A
Question
A block rests on a horizontal frictionless table. It is attached to a spring and set into motion. Consider what will happen to the frequency or period in each of the following situations.
Increase, decrease, or stay the same are the answer choices!
If the amplitude of the motion is cut in half, the period will ______.
If the spring constant is doubled (stiffer spring), the frequency will _______.
If the spring constant is doubled (stiffer spring), the period will _______.
If the mass of the block is cut in half, the frequency will _______.
Explanation / Answer
T = 2*pi*sqrt (m/k)
A.
period does not depend on amplitude so
period will not change
B.
f = 1/T
T is inversly proportional to sqrt (k), So frequency will be directly proportional to sqrt(k)
f2/f1 = sqrt (k2/k1)
f2 = f1*sqrt (2*k1/k1)
f2 = f1*sqrt (2) = 1.414*f1
means frequency will increase.
C.
T is inversly proportional to sqrt (k)
T2/T1 = sqrt (k1/k2)
T2 = T1*sqrt (k1/2*k1)
T2 = T1/sqrt (2)
T2 = 0.707*T1
means period will decrease.
D.
T is directly proportional to mass, So frequency will be inversly proportional to mass
f2 = f1*sqrt (m1/m2)
f2 = f1*sqrt (m1/(m1/2))
f2 = f1*sqrt (2) = 1.414*f1
means frequency will increase.
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