Suppose a spring with spring constant 9 N/m is horizontal and has one end attached to a wall and the other end attached to a mass. You want to use the spring to weigh items. You put the spring into motion and find the frequency to be 0.9 Hz (cycles per second). What is the mass? Assume there is no friction.

Assuming no friction present, the only force acting on the mass (in the horizontal direction) is the spring force.
This force obeys the Hooke's Law, so we can write:

[tex]F = -k*x[/tex]

[tex]F = m*a[/tex]

When we have a movement when acceleration is inversely proportional to the displacement, we have a SHM, with an angular frequency defined as follows:

[tex]\omega = \sqrt{\frac{k}{m} } = 2*\pi *f = \sqrt{\frac{k}{m}}[/tex]

[tex]4*\pi ^{2} *f^{2} = \frac{k}{m}[/tex]

[tex]m = \frac{k}{4*\pi ^{2}*f^{2} } =\frac{9N/m}{4*\pi^{2}*(0.9(1/s))^{2} } = 0.3 kg[/tex]

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-03-25T07:08:09+01:00

The mass attached to the spring is 3.55 kg

The restoring force due to the spring is given by:

F = -kx

where k is the spring constant, and

x is the displacement of the spring

Now, the angular frequency of the spring is given by:

[tex]\omega=\sqrt{\frac{k}{m} }[/tex]

where m is the mass of the object attached to the spring

So, frequency:

f = ω/2π = [tex]\frac{1}{2\pi} \sqrt{\frac{k}{m} }[/tex]

m = 4π²f² / k

given that f = 0.9 Hz, and

k = 9 N/m

m = (4×π²×0.9²)/9

m = 3.55 kg

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