A 0.420-kg object attached to a spring with a force constant of 8.00 N/m vibrate
ID: 1292357 • Letter: A
Question
A 0.420-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.8 cm. (a) Calculate the maximum value of its speed. cm/s (b) Calculate the maximum value of its acceleration. cm/s2 (c) Calculate the value of its speed when the object is 9.80 cm from the equilibrium position. cm/s (d) Calculate the value of its acceleration when the object is 9.80 cm from the equilibrium position. cm/s2 (e) Calculate the time interval required for the object to move from x = 0 to x = 3.80 cm. s
Explanation / Answer
A) 0.5*K*A^2 = 0.*M*Vmax^2
Vmax = A*sqrt(K/m)
Vmax = 0.118*sqrt(8/.42)
Vmax = 0.515 m/s
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b) a = -w^2*A = -(K/m)*A
a_max = -2.25 m/s^2
c) 0.5*m*v^2 + 0.5*k*x^2 = 0.5*K*A^2
v^2 = K*(A^2 - x^2)
V^2 = 8*((0.118*0.118)-(0.098*0.098))
v = 0.186 m/s
d) a = -w^2*x = -1.87 m/s^2
e) x = A*sin (wt)
3.8 = 11.8*sin19t
t = 0.172 s
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