Question

A mass of 187 g is attached to a spring and set into simple harmonic motion with a period of 0.286 s. If the total energy of the oscillating system is 6.94 J, determine the following.

(a) maximum speed of the object

m/s

(b) force constant

N/m

(c) amplitude of the motion

m

Answer #1

m=187g=.187 kg. T=0.286 ,E=6.94J

(a) Mechanical energy will always conserved of the system.The Kinetic energy of the spring when it is mean position it is also it's total energy as the potential energy is zero here.

E=1/2 mv^{2}=1/2*.187*v^{2}

6.94=1/2*.187*v^{2}

v^{2}=6.94*2/.187=13.88/.187=74.22

v=?74.22=8.6151m/s

(b)Time period for spring, T=2*pi*?m/k

0.286=2*3.14*?.187/k

?.187/k=0.286/6.28=0.0454

?.187/k=0.0454

Squaring both side, we get

.187/k=0.045*0.045

K=.187/0.045*0.045=.187/.002025=92.34

k=92.34 N/m

C) At the maximum displacement x=A,the speed is zero and hence
kinetic energy is zero. Then, P. E is 1/2 kA^{2}

Potential energy,
U=1/2*k*A^{2}

^{6.94
J=1/2*92.34*A*A}

A^{2}=6.94*2/92.34=0.1503

A=0.38768 m=.388 m

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