CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

SHM in Vertical Spring Block

27.Two masses...

Question

27.Two masses M1 equals to 1 kg and m2 is equal 0.5 kg are suspended Together by a massless spring of spring constant 12.5 Newton per metre when masses are in equilibrium m1 is removed without disturbing the system new amplitude of the oscillation will be 1)30cm 2)50cm 3)80cm 4)60cm

Open in App

Solution

Suggest Corrections

thumbs-up

0

Similar questions

m_{1} = 1 kg

m_{2} = 0.5 kg

are suspended together by a massless spring of spring constant

12.5 N/m

. When masses are in equilibrium,

m_{1}

is removed without disturbing the system. New amplitude of oscillation will be

m_{1} = 1 kg

m_{2} = 0.5 kg

are suspended together by a massless spring of spring constant

12.5 N/m

. When masses are in equilibrium,

m_{1}

is removed without disturbing the system. New amplitude of oscillation will be

m_{1}

m_{2}

are suspended together by a massless spring of spring constant

k

(Fig). When the masses are in equilibrium,

m_{1}

is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of

m_{2}

.

m_{1}

m_{2}

are suspended together by a mass less spring of constant K. given the mass are in equilibrium

m_{2}

is removed without disturbing the system,the amplitude of oscillation will be

m_{1}

m_{2}

are suspended together by a massless spring of spring constant

k

. When the masses are in equilibrium,

m_{2}

is removed without disturbing the system. The amplitude of SHM is

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Applying SHM

PHYSICS

Watch in App

explore_more_icon

Explore more

SHM in Vertical Spring Block

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon