91影视

In this problem, we are given the equation of motion as \(x(t) = 0.5 \cos (5 t + \pi/4)\). This is the equation for simple harmonic motion in the form of \(x(t) = A \cos (\omega t + \phi)\), where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase angle. By comparing the given equation, we can identify the following parameters: Amplitude (A): \(0.5 \mathrm{~m}\) Angular Frequency (\(\omega\)): \(5 \mathrm{~s^{-1}}\) Phase Angle (\(\phi\)): \(\pi/4 \mathrm{~rad}\)

Next, we will find the frequency (f) and the period (T) of the oscillation, since they will be useful in determining the spring constant. We can use the following relationships between angular frequency, frequency, and period: \(\omega = 2 \pi f\) and \(T = \dfrac{1}{f}\) From the given equation, we have an angular frequency (\(\omega\)) of 5 : \(f = \dfrac{\omega}{2 \pi} = \dfrac{5}{2 \pi} \approx 0.796 \mathrm{~Hz}\) Now we can calculate the period (T): \(T = \dfrac{1}{f} = \dfrac{1}{0.796} \approx 1.256 \mathrm{~s}\)

To find the spring constant, we will use Hooke's Law and the relationship between the period and mass on a spring: Hooke's Law: \(F = -kx\) Relationship between period and mass: \(T = 2\pi \sqrt{\frac{m}{k}}\) The force acting on the mass due to the spring (F) is the product of mass (m) and acceleration (a). In simple harmonic motion, acceleration is given by \(a = -\omega^{2} x\). Thus, \(F = -ma = m\omega^{2}x\) Substituting Hooke's Law into this equation, we get: \(kx = m\omega^{2}x\) Now, we will solve for the spring constant (k). Since x 鈮� 0, we can divide both sides by x: \(k = m\omega^{2}\) Finally, we will plug in the given mass (m = 5.00 kg) and the angular frequency (\(\omega\) = 5 s鈦宦�): \(k = (5.00 \mathrm{~kg}) (5 \mathrm{~s^{-1}})^{2} = 5 \cdot 25 = 125 \mathrm{~N/m}\) The spring constant is 125 N/m.