Question

an object of mass 1 g is hung from a spring and set in oscillatory motion .At t=0 the displacement is 43.75cm and the acceleration is -1.754cm/sec.what is the spring constant?​

Answer 1

The spring constant is $4.0\cdot 10^{-5} N/m$

Explanation:

For an object in a simple harmonic motion, the acceleration of the object is related to the displacement by

$a=-\omega^2 x$

where

a is the acceleration

$\omega$ is the angular frequency

x is the displacement

The angular frequency is defined as

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

where

k is the spring constant

m is the mass

Substituting the second equation into the first one, we get

$a=-\frac{k}{m}x$

In this problem we have

m = 1 g = 0.001 kg

And at t=0,

x = 43.75 cm

a = -1.754 cm/s

Therefore, we can re-arrange the equation above to find the spring constant:

$k=-\frac{ma}{x}=-\frac{(0.001)(-1.754)}{43.75}=4.0\cdot 10^{-5} N/m$

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