1- The period of a simple pendulum is 4.2 s at a point where g = 9.81 m/s 2 . Wh
ID: 1405839 • Letter: 1
Question
1- The period of a simple pendulum is 4.2 s at a point where g = 9.81 m/s2. What would be the period of this pendulum if it were on the Moon, where the acceleration due to gravity is one-sixth that on Earth?
.......... s.
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2- A 2.0 kg object oscillates on a spring with an amplitude of 9.5 cm. Its maximum acceleration is 3.00 m/s2. Find the total energy.
........ J.
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3- An 85 kg person steps into a car of mass 2700 kg, causing it to sink 2.35 cm on its springs. Assuming no damping, with what frequency will the car and passenger vibrate on the springs?
.......... Hz.
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4- Military specifications often call for electronic devices to be able to withstand accelerations of 10g = 98.1 m/s2. To make sure that their products meet this specification, manufacturers test them using a shaking table that can vibrate a device at various specified frequencies and amplitudes. If a device is given a vibration of amplitude 2.2 cm, what should its frequency be in order to test for compliance with the 10g military specification?
........... Hz.
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5- Two systems each consist of a spring with one end attached to a block and the other end attached to a wall. The identical springs are horizontal, and the blocks are supported from below by a frictionless horizontal table. The blocks are oscillating in simple harmonic motions with equal amplitudes. However, the mass of block A is four times as large as the mass of block B. How do their maximum speeds compare?
.. vAmax = vBmax
.. vAmax = 2vBmax
.. vAmax = 1/2vBmax
... This comparison cannot be done by using the data given.
Explanation / Answer
1)
timeperiod = T = 2*pi*sqrt(L/g)
T2/T1 = sqrt(g1/g2)
given g2 = g1/6
T2 = T1*sqrt(6)
T2 = 4.2*sqrt6
T2 = 10.3 s <<<---answer
++++++++
2)
total energy = KE + PE
TE = 0.5*m*w^2*(A^2-x^2) + 0.5*k*x^2
TE = 0.5*m*w^2*A^2
maximum acceleration a = w^2*A
TE = 0.5*m*a*A
TE = 0.5*2*3*0.095
TE = 0.285 J <<<<---answer
3)
F = k*x
K = F/x = mg/x = (85*9.8)/0.0235
K = 35446.8 N/m
w = sqrt(k/(M+m))
w = sqrt(35446.8/(2700+85))
w = 3.57
f = w/2pi
w = 0.57 Hz <<<_answer
+++
4)
in SHM
w^2 = a/A
w = sqrt(a/A)
2*pi*f = sqrt(a/A)
f = (1/2pi)*Sqrt(10g/A)
f = (1/6.28)*Sqrt(98.1/0.022)
f = 10.63 Hz
5)
EA = 0.5*KA*A^2
EB = 0.5*KB*A^2
KA = KB
EA = EB
0.5*mA*VAmax^2 = 0.5*mB*VBmax^2
mA = 4*mB
vBmax^2 = 4*VAmax^2
vBmax = 2*vAmax
vAmax = 1/2*vBmax <<<---answer
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