Question

-w'r, (25%) Problem 4: Any system for which the acceleration is linearly proportional to the position with a negative proportionality constant), or a = undergoes simple harmonic motion, a form of oscillatory motion. The mathematical solution to this is (t) = A coswt) where A is the amplitude and w=2nf = 2 is the angular frequency (fis the frequency in Hz and Tis the period). For a mass on a spring, w2 = A 0.1 kg mass on a spring with k = 5 N/m is stretched from its equilibrium position by 15 cm and then released. 4 8% Part (a) Sketch the position versus time for this oscillator for 4 seconds. Be as quantitative and precise as possible and be sure to label your graph. Mark an X where all of the energy in the oscillator is kinetic energy. Mark an O where all the energy of the oscillator is elastic potential energy. A8% Part (b) Find the amplitude (4) in centimeters. 8% Part (c) Find the period (7) in seconds. 8% Part (d) Find the frequency (7 in Hz. 8% Part (e) Find the angular frequency (w) in rad/s. 8% Part (1) Find the maximum speed (max) in m/s. 8% Part (9) Find the maximum acceleration (amax) in m/s2 8% Part (h) Find the total energy (E) in joules. 8% Part (0) What is the mass's speed when it is at x = 10 cm (i.e., not at the maximum stretching)? Give your answer in m/s. A 8% Part (i) If you initially stretched the spring by 30 cm instead of 15 cm, what would the oscillation frequency be in Hz? 4 8% Part (k) If you initially stretched the spring by 30 cm instead of 15 cm, what would the oscillation period be in seconds? 8% Part (1) If the spring had twice the spring constant, what would be the new frequency of the oscillations in Hz? 8% Part (m) If the object on the spring was four times as massive, what would be the frequency of the oscillations?

There's a lot going on here and I am overwhelmed. I have no idea how to start this.

Answer 1

Lets start from the easy part.

b. They say its being streched by 15 cm from equilibrium so our Amplitude is 15 cm

c. If you rerrange the equations given you will get T = 2 pi sqrt (m/k)

so T= 2 pi sqrt (0.1/5 ) =  0.89 s

d. f= 1/T = 1.125 Hz

e. w= 2 pi f = 7.07 rad/s

f. Vmax= Aw = 0.15 *7.07 = 1.06 m/s

g.amax= w^2 A= 7.07*7.07*0.15 = 7.50 m/s2

h.Total energy= Max Kinetic energy= 0.5 m vmax^2 = 0.5*0.1*1.06*1.06 =0.056 J

i. The velocity equation is after differentiating the displacement equation

v= - Aw sin(wt) , now you need t when x=10 cm

use x= A cos(wt)

10= 15 cos(7.07t)

t= 6.82 s

so v= 0.15*7.07 sin(7.07*6.82) = 0.79 m/s

j. Remains same, Frequency is independent of the amplitude

k. remains same

l. f= 1/2pi (sqrt(k/m))

so if k is double

f will increase by sqrt2

so new f= sqrt2 *1.125 = 2.250 Hz

m. If the mass is four times, the f is 1/2

so new f = 1.125/2 = 0.5625 Hz

Now for the first part you draw a cosine graph with time on x axis and displacement on y axis.

Use the time period and divide it by four to mark each peak and zero , now they want for 4 seconds and time period is 0.89 so you should have exactly 4.5 full waves while drawing.

Now Kinetic energy is maximum when displacement is zero so mark x at any of the one points where graph cuts the time axis and mark o at any of the peaks.