A 1 kg body is attacked to a spring of negligible mass. The spring vibrates in s

Question

A 1 kg body is attacked to a spring of negligible mass. The spring vibrates in simple harmonic motion according to the equation x = 0.15 cos (pi t + pi/8) where x is in meters and t is in seconds. What is the amplitude of its motion (in meters)? What is its period T? What is its frequency f? What is the phase constant? The total number of vibrations this body makes in 10 s is What is the spring constant? What is the total energy of the oscillator? FORMULAS X = A cos (omega t + phi) T = 1/f T = 2 pi squareroot m/k E = 1/2KA^2 omega 2 pi/T

Explanation / Answer

Given
mass of body is m = 1 kg

equation is x = 0.15cos(pit+pi/8)

which is in the form of X = A cos (wt+phi)

wher A is amplitude , W is angular frequecncy , phi is phase angle

  
1) amplitude A = 0.15 m

2) time period T = 1/f, and W = 2pif ==> f = W/2pi = pi/2pi = 1/2 Hz

       T = 1/(1/2) = 2 s

3) frequency f = 0.5 Hz

4)

   phase constant phi = pi/8

5) the no of vibrations made by it in 10 s is , T = 2 s, then it can make 5 vibrations in 10 s

6) spring constant k =?

   T = 2pi sqrt(m/k) ==> k = 4pi^2 *m/T^2 = 4pi^2*1/2^2 = 9.86960 N/m

7) total energy of the oscillatior

   E = 1/2 k A^2 = 0.5*9.86960*0.15^2 J = 0.111033 J

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