Explanation
vmax=Aω=0.866msAt x=A2;v=ω√A2−x2v=√32ωA=√32vmaxv=√32×0.866=0.75ms
for a parallel combination k=k1+k2T=2π√mk
=2π√m(k1+k2)
T1=2π√mk1
⇒k1=(2π)2mT21
k2=(2π)2mT22
k1+k2=(2π)2(mT21+mT22)
∴T1T2√T1+T2=T
vmax=15 cm/s;T=0.628 s=2π10sω=2πT=10 rad/svmax=AωA=1.5 cm
V=ω√A2−x2
V1=ω√A2−x2
V1=ω√A2−(0.6)2
V2=ω√A2−x2
V2=ω√A2−(0.8)2
V1V2=√(12−(0.6)2√(12−(0.8)2
4V2=0.80.6
V2=3 cm/sec
x=4sin(80t+π2)ω1=80;2πT1=80;T1=π40secy=2cos(60t+π3)ω2=60;2πT2=60;T2=π30secT1T2=34
Let y is the displacement y=asinωt
ω=2πT
It is given that,
Displacement y = 1
Amplitude a = 2
So 1=3sin2πtT
Since, acceleration in SHM is a=ω2y
And velocity in SHM is v=ω√a2−y2....(1)
∵v=ω
Put all the values in equation (1)
4π2T2=2πT(√a2−1)
2πT=(√3)
T=2π√3
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