Question #171383

Q3: A spring is mounted linear ,by attaching a spring balanced to

the free end and pulling sideways, we determine that the force of

(4 N ) causes a displacement of 0.02 m .we attach a 2 kg body to

the end ,pull it a side distance of 0.04 m and release it .Find :

1) the force constant of the spring ?

2) the period and frequency of vibration ?

3) compute the maximum velocity attained by the vibrating

body ?


1
Expert's answer
2021-03-16T08:32:52-0400

1) F = force applied to pull the spring = 4 N

x=stretch of the spring=0.02mx = stretch \space of\space the\space spring = 0.02 m

k=spring constantk = spring \space constant

Using the equation

F=kxF = k x

4=k(0.02)4 = k (0.02)

k=40.02=200N/mk = \frac{4}{0.02}=200N/m


2) m = mass of glider = 2.0 kg

Angular frequency is given as w=km=2002=100=10rad/sw = \sqrt{\frac{k}{m}}= \sqrt{\frac{200}{2}}= \sqrt{100}=10rad/s

Frequency is given as f=w2π=102π=1.5915Hzf=\frac{w}{2\pi}=\frac{10}{2\pi}=1.5915 Hz

Period, T=1f=11.5915=0.6283T=\frac{1}{f}=\frac{1}{1.5915}=0.6283


3) At the extreme position, the mass will have only PE = 0.5kx2=0.5(200)(0.04)2=0.16J0.5 kx^2 = 0.5 ( 200)( 0.04)^2 = 0.16 J

At the mid position, ( x = 0.02 m), the body will have both KE and PE

0.16J=0.5(200)(0.02)2+0.5(2)(v)20.16 J = 0.5 ( 200)(0.02)^2 + 0.5 ( 2)( v)^2

0.16=0.04+v20.16 = 0.04 + v^2

V=0.346m/sV = 0.346 m/s




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