A block of mass m = 0.5 kg is attached to a horizontal spring
whose spring constant is k = 50 N/m. At t = 0.1s, the position is x
= -0.2 m and the velocity is v_{x} = +0.5 m/s. It is
assumed that x(t) = A sin(t+
φ). (a) Determine the amplitude and phase constant. (b) At what
time does the condition x = 0.2 m and v_{x}= -0.5 m/s first
occur?

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b) Ans:-Now, x(t)=A \sin (\omega t+\phi)Put x=0.2 \mathrm{~m}, w=10, \phi=-77.02. and find out t for x=0.2 \mathrm{~m}.\begin{array}{l}0.2=0.2061 \sin (70 t-77.02) \\0.9704=\sin (20 t-77.02) \\10 t-77.02=76.0253 \\10 t=153.04 \\t=t=15.304 \mathrm{sec}\end{array}similarly, U_{n}(t)=A \omega as (\omega t+\phi)\begin{array}{c}\Rightarrow-0.5=0.2061 \times 10 \cos (70 \times 0=79.02) \\10 t-77.02=\cos 7(0.2426) \\10 t-77.02=75.95 \\\Rightarrow t=15.29 \mathrm{scc} \Rightarrow v_{n}=-0.5 \frac{\mathrm{m}^{3 c}}{} \text { zगt occur }^{11}\end{array}\Rightarrow{ }^{\prime \prime} U_{n}=-0.5 \frac{\mathrm{m}}{\mathrm{sic}} गt occus" ...