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Answer :
Following are the solution to the given points:
For part A)
The force exerted by the spring can be calculated using Hooke's law as follows:
[tex]\to F_s = -k\times x\\\\[/tex]
[tex]\to\text{Spring constant} (K) = 94.6 \frac{N}{m}\\\\\to x = \frac{\text{stretch}}{\text{extension of spring}} = 0.173\ m\\\\[/tex]
As a consequence, the season's force is as follows:
[tex]\to F_s = -94.6\times 0.173 = -16.3658 \ N[/tex]
(The -ve symbol denotes that the force exerted by the spring is in the opposite direction).
Calculating the three significant figures:
[tex]\to F_s = -16.4 \ N\\\\[/tex]
For part B)
The angular frequency of oscillatory motion is given by:
[tex]\to w = \sqrt{ (\frac{k}{m})}\\\\[/tex]
[tex]= \sqrt{ (\frac{94.6}{0.792})} \\\\= 10.929 \ \frac{rad}{sec}\\\\[/tex]
Calculating the three significant figures:
[tex]\to w = 10.9 \ \frac{rad}{sec}[/tex]
For part C)
maximum speed in SHM movement is provided by:
[tex]\to V_{max} = A \times w \\\\\to \text{Amplitude}\ (A) = 0.173 \ m\\\\[/tex]
So,
[tex]\to V_{max} = 0.173\times 10.929\\\\[/tex]
[tex]= 1.89 \ \frac{m}{s}\\\\[/tex]
For part D)
The maximum acceleration of the block is given by:
[tex]\to a_{max} = A\times w^2\\\\[/tex]
[tex]= 0.173\times 10.929^2\\\\ = 20.6636 \ \frac{m}{s^2}\\\\[/tex]
Calculating the three significant figures:
[tex]\to a_{max} = 20.7\ \frac{m}{s^2}\\\\[/tex]
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Answer:
Explanation:
a) F = -kx = -99.9 * 0.12 = -11.988N
b)
[tex]F = ma = -kx\\ a = -\frac{k}{m} x\\[/tex]
Using for x,v,a :
[tex]x = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\a = -\omega ^{2} A cos(\omega t + \phi)\\a= - \omega ^2 x[/tex]
gives the angular frequency:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
ω = 12.6 1/s
c) v is max, when [tex]sin(\omega t + \phi) = 1[/tex]:
[tex]|v_{max}| = |-A\omega| = 0.12*12.6 = 1.512 \frac{m}{s}[/tex]
d) a is max, when [tex]cos(\omega t + \phi) = 1[/tex]:
[tex]|a_{max}|= |-\omega^2 A| = 12.6^2 * 0.12 = 19.05N[/tex]
