Question #184431

a body of mass 80 g is attached to the end of a helical spring with a spring constant of 4500 dynes/cm, and is made to vibrate with an amplitude of 16 cm. calculate the period of vibration, the maximum velocity of the body, and the velocity of the body when it is 10 cm from the equilibrium position


Expert's answer

m=80 g=80×10−3 kgm=80\space g=80\times10^{-3}\space kg

Spring Constant, k=4500 dynes/cmk=4500\space dynes/cm=4500×10−3 N/m=4500\times10^{-3}\space N/m

Amplitude, A=16 cm=16×10−2 mA=16\space cm=16\times10^{-2}\space m

Time Period, T=2πmkT=2\pi mk

T=2×3.14×(80×10−3)×(4500×10−3)=2.261 sT=2\times3.14\times(80\times10^{-3})\times(4500\times10^{-3})=2.261\space s


Maximum Velocity, vmax=Akm=0.0576 m/sv_{max}=Akm=0.0576\space m/s


Velocity at x=10cm,

v=km(A−x)=0.0216 m/sv=km(A-x)=0.0216\space m/s


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