As an example of simple harmonic motion, we first consider the motion of a block of mass (m) that can slide without friction along a horizontal surface. The mass is attached to a spring with spring constant (k) which is attached to a wall on the other end. We introduce a one-dimensional coordinate system to describe the position of the mass, such that the (x) axis is co-linear with the motion, the origin is located where the spring is at rest, and the positive direction corresponds to the spring being extended. This “spring-mass system” is illustrated in Figure (PageIndex{1}).
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: A horizontal spring-mass system oscillating about the origin with an amplitude (A).</p>
<p>We assume that the force exerted by the spring on the mass is given by Hooke’s Law: <egin{aligned} vec F = -kx hat xend{aligned}> where (x) is the position of the mass. The only other forces exerted on the mass are its weight and the normal force from the horizontal surface, which are equal in magnitude and opposite in direction. Therefore, the net force on the mass is the force from the spring.</p>
<p>As we saw in Section 8.4, if the spring is compressed (or extended) by a distance (A) relative to the rest position, and the mass is then released, the mass will oscillate back and forth between (x=pm A)1, which is illustrated in Figure (PageIndex{1}). We call (A) the “amplitude of the motion”. When the mass is at (x=pm A), its speed is zero, as these points correspond to the location where the mass “turns around”.</p>
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Exercise (PageIndex{2})
What is the SI unit for angular frequency?
( ext{Hz}) ( ext{rad/s}) ( ext{N}^{1/2} ext{m}^{-1/2} ext{kg}^{-1/2}) All of the above Answer
All of the above
Olivia”s Thoughts
In Chapter3, we found, (x(t)), from a function, (a(t)), by using simple integration. You may be wondering why we can’t do the same thing in order to find (x(t)) for the mass-spring system. The difference is that, before, the acceleration was a function of time. Here, the acceleration is a function of (x). This means that we have to use a different method to solve for (x(t)), which is why we are making these “guesses” to solve a differential equation.
We still need to identify what the constants (A) and (phi) have to do with the motion of the mass. The constant (A) is the maximal value that (x(t)) can take (when the cosine is equal to 1). This corresponds to the amplitude of the motion of the mass, which we already had labeled, (A). The constant, (phi), is called the “phase” and depends on when we choose (t=0) to be. Suppose that we define time (t=0) to be when the mass is at (x=A); in that case: <egin{aligned} x(t=0) &= A\ A cos(omega t + phi) &= A\ A cos(omega (0) + phi) &= A\ cos(phi) &= 1\ herefore phi = 0end{aligned}> If we define (t=0) to be when the mass is at (x=A), then the phase, (phi), is zero. In general, the value of (phi) can take any value between (-pi) and (+pi)3 and, jonathanlewisforcongress.comically, corresponds to our choice of when (t=0) (i.e. the position of the mass when we choose (t=0)).
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Since we have determined the position as a function of time for the mass, its velocity and acceleration as a function of time are easily found by taking the corresponding time derivatives: <egin{aligned} x(t) &= A cos(omega t + phi)\ v(t) &= frac{d}{dt}x(t) = -Aomegasin(omega t + phi)\ a(t)&= frac{d}{dt}v(t) = -Aomega^2cos(omega t + phi)end{aligned}>
Exercise (PageIndex{3})
What is the value of (phi) if we choose (t=0) to be when the mass is at (x=0) and moving in the positive (x) direction?
(pi) (-pi) (pi/2) (-pi/2) Answer
The position of the mass is described by a sinusoidal function of time; we call this type of motion “simple harmonic motion”. The position and velocity as a function of time for a spring-mass system with (m=1 ext{kg}), (k=4 ext{N/m}), (A=10 ext{m}) are shown in Figure (PageIndex{2}) for two different choices of the phase, (phi=0) and (phi=pi/2).
