Question

Q (3). A mass m is suspended from a spring system shown below. Obtain an equivalent spring mass system
with single spring and mass. Given that k2 = k3= 250 N/m and Ki = 500 N/m.
WHO​

Answer 1

sorry

m dt

m dt 2

m dt 2

m dt 2 d

m dt 2 d 2

m dt 2 d 2 x

m dt 2 d 2 x

m dt 2 d 2 x +b

m dt 2 d 2 x +b dt

m dt 2 d 2 x +b dtdx

m dt 2 d 2 x +b dtdx

m dt 2 d 2 x +b dtdx +kx=0−−−(1)

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ=

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c±

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c 2

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c 2 −4mk=0

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c 2 −4mk=0Given m=0.5 kg

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c 2 −4mk=0Given m=0.5 kgk=200 N/m

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c 2 −4mk=0Given m=0.5 kgk=200 N/mC=

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c 2 −4mk=0Given m=0.5 kgk=200 N/mC= 4 mk

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c 2 −4mk=0Given m=0.5 kgk=200 N/mC= 4 mk

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c 2 −4mk=0Given m=0.5 kgk=200 N/mC= 4 mk =

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c 2 −4mk=0Given m=0.5 kgk=200 N/mC= 4 mk = 400

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c 2 −4mk=0Given m=0.5 kgk=200 N/mC= 4 mk = 400

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c 2 −4mk=0Given m=0.5 kgk=200 N/mC= 4 mk = 400 =20 kg/s

m dt 2 d 2 x +b dtdx +kx=0−−−(1)Solution is of the form x=e λt Putting x=e λt in (1)mλ 2 +bλ+k=0λ= 2m−c± c 2 −4mk As critical dumping, c 2 −4mk=0Given m=0.5 kgk=200 N/mC= 4 mk = 400 =20 kg/sDamping coefficient =20 kg/s