Q1. A body of mass 4.35 kg moves along the x -axis, and its motion is described
ID: 2103097 • Letter: Q
Question
Q1. A body of mass 4.35 kg moves along the x-axis, and its motion is described by the equation x = 6 sin (23 t + 2),where x is in m and t is in s.
A) Find the resultant force on the body at t = 0.100 s.
B) Find the resultant force on the body at t = 0.200 s.
Q2. A small mass m is supported by a vertical spring. When an additional 102 g is attached to the original mass, the system begins to oscillate at a frequency of 0.341 Hz. When the oscillations die out, the spring is found to have increased in length by 11.3 cm. Find m.
Explanation / Answer
FOR QUESTION 1; acceleration ;a =d2x/dt^2 = 6*23^2*sin (23 t + 2) putting value of t at t=0.1 a=2907.91 m/s^2 F=m*a=12.649 kN at t=0.2 a=988.83 m/s^2 F= 4.3kN FOR question 2: k = (m+0.102)w^2 ; here w= angular frequency = 2*pi*0.341 Hz = 2.143 hence k = (m+0.102)*4.59 originally kx = mg; now k(x+0.113) = (m+0.102)g equating we get; k(0.113) = 0.102g (m+0.102)*4.59*(0.113) =0.102g solving we get m = 1.825 kg
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