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2 years ago

Help please in science

Physics

1 answer:

slamgirl [31]2 years ago

3 0

The amplitude is 0.8 cm. The wavelength is 2.0 cm.

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An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the

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Answer:

The value is $A = 0.014 \ m$

Explanation:

From the question we are told that

The mass of the object is $m = 2.0 \ kg$

The unstressed length of the string is $l = 0.08 \ m$

The length of the spring when it is at equilibrium is $l_e = 5.9 \ cm = 0.059 \ m$

The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed of the spring is mathematically represented as

$u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

and $w$ is the angular frequency which is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

$u = A * \sqrt{\frac{k}{m} }$

=> $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

$b = l -l_e$

=> $b = 0.08 - 0.05 9$

=> $b = 0.021 \ m$

Generally at equilibrium position the net force acting on the spring is

$k * b - mg = 0$

=> $k * 0.021 - 2 * 9.8 = 0$

=> $k = 933 \ N/m$

$A = 0.30 * \sqrt{\frac{2}{933} }$

=> $A = 0.014 \ m$

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