A particle having mass 10 g oscillates according to the equation `x=(2.0cm) sin[(100s^-1)t+pi/6]`. Find the amplitude the time period and the spring constant

Answer

A, B, C

Explanation

`x=(2.0cm)sin[(100s^-1)t+pi/8]`
m=10g`
a. Amplitude =2cm
`omega=100sec^-1`
`:. T=(2pi)/100=pi/50sec`
`=0.063sec.
We know that
`T=2pi m/k`
`rarr pi^2=4pi^2m/k`
`rarr k=(4pi^2)/T^2=10.5dyne/cm`
`=100N/m`
`[Because 0=(2pi)/T=100sec^-1]`
b. At `t=0, x=2 cm sin pi/6`
`=2xx1/2=1cm` from the mean position
`We know that
`x=Asin(omegat+pi)`,
`v=aomegacos(omegat+phi)`
`=2xx100cos(0+pi/6)`
=200xxsqrt3/2`
`=100sqrt3cmsec^-1`
`=1.73ms^-1`
c. `a=-w^2x`
`=100^2xx1=1cm/s^2`