The period of oscillation of an object in an ideal spring and mass system is 0.50 s and the amplitude is 5 cm. Determine the velocity of the body at the equilibrium point
Solution.
T=0.50s;T=0.50s;T=0.50s;
xmax=5cm=0.05m;x_{max}=5cm=0.05m;xmax=5cm=0.05m;
vmax−?;v_{max}-?;vmax−?;
vmax=ωxmax;v_{max}=\omega x_{max};vmax=ωxmax;
ω=2πT;\omega = \dfrac{2\pi}{T};ω=T2π;
ω=2⋅3.140.50s=12.56s−1;\omega=\dfrac{2\sdot3.14 }{0.50s}=12.56s^{-1};ω=0.50s2⋅3.14=12.56s−1;
vmax=12.56s−1⋅0.05m=0.628m/s;v_{max}=12.56s^{-1}\sdot0.05m=0.628m/s;vmax=12.56s−1⋅0.05m=0.628m/s;
Answer: vmax=0.628m/s.v_{max}=0.628m/s.vmax=0.628m/s.
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