A simple harmonic oscillator consists of a 1.1 kg block attached to a spring (k

Question

A simple harmonic oscillator consists of a 1.1 kg block attached to a spring (k = 220 N/m). The block slides on a horizontal frictionless surface about the equilibrium point x = 0 with a total mechanical energy of 5.0 J. (a) What is the amplitude of the oscillation? (b) How many oscillations does the block complete in 16 s? (c) What is the maximum kinetic energy attained by the block? (d) What is the speed of the block at x = 0.17 m? (a) Number Units (b) Number Units (c) Number Units (d) Number Units

Explanation / Answer

E = Total Mechanical Energy

(E = The Kinetic Energy plus the Potential Energy at any given point. If it is at equilibrium position, then it is all KE and no PE, but if x = A, then it is all PE and no KE because the Velocity is zero, For any other position, the E has some KE and some PE, but the still add up to E)

E = (1/2) k A^2

A is the amplitude and k is is the spring constant.
A = sqrt(2E/k)
A = sqrt ( 2 * 5.0 J / 220 N/m)
A = 0.2132 m
b)
f = (1/2pi) * sqrt(k/m)
T = 16 sec
N/T = f
N = f * T
N = (T/2pi) * sqrt(k/m)
N = (16 seconds/ 2*pi) * sqrt(220 N/m / 1.1 kg)
N = 36.012oscillations

C) E = KE + PE
The most KE we can get is at the center position where PE = 0 J.
So KE = E - PE = E
KE = 5.0 J
D)
KE = E - PE
and we know that the PE of a spring is (1/2) kx^2
and we know that KE = (1/2) mv^2, so .....
(1/2) mv^2 = E - ((1/2) kx^2)
v = sqrt[(2/m) * (E - ((1/2) kx^2))] = sqrt[(2/1.1) * (5 - ((1/2) 220*0.17^2))] =2.45 m/s



v = sqrt[(2E - kx^2) / m]

There, just put in your numbers

v = sqrt{(2*6.0 J - (200 N/m)(0.15 m)^2) / 0.90 kg]

v = 8.33 m/s

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