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## _-Physics 111 - Iowa State University

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Physics 111
Oscillations
An object attached to a spring exhibits simple harmonic motion with an amplitude of 4.0 cm. When the object is 2.0 cm from the equilibrium position, what percentage of its total mechanical energy is in the form of potential energy?(a) One-quarter.(b) One-third.(c) One-half.(d) Two-thirds.(e) Three-quarters.
Concept Checker:
Express the ratio of the potential energy of the object when it is 2.0 cm from the equilibrium position to its total energy:U(x) = .5Kx^2 = x^2= 2^2 = 4E total = 5KA^2= A^2 = 4^2 = 164/16 = ¼ the total energy
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?(a)Vamax =Vbmax(b)Vamax = 2Vb max(c)Vamax = .5Vbmax(d) This comparison cannot be done by using the data given.
Concept Checker 2:
Vamax = (K / m * A) ^1/2Vbmax = (K / 4m * A) ^1/2Since A and K are constant, can make a portion whereVamax/Vbmax = (m / 4 M) ^1/2Vb= ½
SOLUTION: C, .5
Two simple pendulums are related as follows. Pendulum A has a length LA and a bob of massmA; pendulum B has a length LB and a bob of massmB. If the frequency of A is one-third that of B, then(a) LA = 3LB andmA= 3mB,(b) LA = 9LB andmA=mB,(c) LA = 9LB, regardless of the ratiomA/mB,(d) A = (3LB)^1/2, regardless of the ratiomA/mB.
Last Conceptual Question:
f= 1/ 2pie* (g/L)^.5 Or L = g / 4 pie^2 f^2La = g / 4 pie^2 f^2Lb = g / 4 pie^2 f^2G is constant, and pie ^2 is constantJust need to compare La, Lb,faandfbSo La/Lb = f^2 / (1/3)fb^2 = 9Answer C
A mass of 2 kg is attached to a spring with constant 18 N/m. It is then displaced to the pointx= 2 . How much time does it take for the block to travel to the pointx= 1 ?
Practice Problems:
X =Acos(wt)W = 2 pie / T = (k/m)^1/2w= (18/2)^1/2 = 31 = 2cos( 3 t )T = .35 seconds
Solution:
A mass of 2 kg oscillating on a spring with constant 4 N/m passes through its equilibrium point with a velocity of 8 m/s. What is the energy of the system at this point? From your answer derive the maximum displacement,xmof the mass.
Practice Problems
K = .5 mv^2 = .5(2)(8)^2 = 64 joulesU = .5 kx^2Since energy is conserved:K = .5 mv^2 = U = .5 kx^2 = 64X m = (x *2 / k) ^1/2 = 32^1/2 = 5.65 almost
Solution:
One end of alight spring with a spring constant 10 N/m is attached to a vertical support, while a mass is attached to the other end. The mass is pulled down and released, and exhibits simple harmonic motion with a period of .2 pie. The mass is.1 kg.25kg.4kg.04 kg.025kg
Last Problem:
T= 2 pie (m/k)^1/2M = kT^2 / 4 pie^2= (10)(.2 pie)^2 / 4 pie ^2M= .1 kg