Answer:
Part(a): The amplitude of motion is 0.94 m.
Part(b): The maximum acceleration of the block is 165.13 [tex]\bf{m~s^{-2}}[/tex].
Part(c): The maximum force that the spring exerts on the block is 305.5 N.
Explanation:
Part(a):
Given the mass ([tex]m[/tex]) of the block is 1.85 Kg, the force constant ([tex]k[/tex]) is 325 [tex]N~m^{-1}[/tex].Initially the spring is neither stretched nor compressed, which indicates that the block at this situation is in its equilibrium position where the maximum velocity of the block is [tex]v_{max} = 12.5~m~s^{-1}[/tex]. If '[tex]A[/tex]' be the amplitude of motion, then the velocity of the particle executing simple harmonic motion at any instant of position ([tex]x[/tex]) is
[tex]v = \omega~{\sqrt{A^{2} - x^{2}}}......................................................(I)[/tex]
where [tex]\omega[/tex] is the natural angular frequency.
At equilibrium position, x = 0. So, the maximum velocity ([tex]v_{max}[/tex]), using equation (I) can be written as
[tex]&& v_{max} = \omega~\sqrt{A^{2} - 0}\\&or,& v_{max} = \omega \times A\\&or,& A = \dfrac{v_{max}}{\omega} = v_{max} \times \sqrt{\dfrac{m}{k}}\\&or,& A = (12.5)~m~s^{-1} \times \sqrt{\dfrac{1.85~Kg}{325~N~m^{-1}}} = 0.94~m[/tex]
Part(b):
The acceleration ([tex]a[/tex]) of a particle executing SHM is given by
[tex]a = \omega^{2}~x = \dfrac{k}{m}~x................................................(II)[/tex]
The block will gain its maximum acceleration when it is at a distance equal to its amplitude. SO from equation (II), the maximum acceleration ([tex]a_{max}[/tex]) of the block is
[tex]a_{max} = \dfrac{k}{m} \times A = \dfrac{325~n~m^{-1}}{1.85~Kg} \times 0.94~m = 165.13~m~s^{-2}[/tex]
Part(c):
The block will experience a maximum restoring force ([tex]F_{max}[/tex]) when it is at a distance [tex]x = A[/tex]. So, the value of the maximum force is
[tex]F_{max} = k \times A = 325~N~m^{-1} \times 0.94~m = 305.5~N[/tex]
