spring stretches from by 1.8 cm and 2.8 kg block is suspended from its end the mass that has to be attached to the spring so that its frequency of the vibration is three Hz is

Question

TARUN SHARMA · Answer

WHEN mass 2.8 kg attached then the system is in equalibrium with the spring so ; kx = mg k=mg/x and as we know frequency= omega/2pi omega^2=k/m1 m1 total mass by equating values in equation you will get your answer

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spring stretches from by 1.8 cm and 2.8 kg block is suspended from its end the mass that has to be attached to the spring so that its frequency of the vibration is three Hz is

spring stretches from by 1.8 cm and 2.8 kg block is suspended from its end the mass that has to be attached to the spring so that its frequency of the vibration is three Hz is

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spring stretches from by 1.8 cm and 2.8 kg block is suspended from its end the mass that has to be attached to the spring so that its frequency of the vibration is three Hz is

spring stretches from by 1.8 cm and 2.8 kg block is suspended from its end the mass that has to be attached to the spring so that its frequency of the vibration is three Hz is

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