Physics, 23.07.2019 00:30 leenzazou98241

A0.50-kg object is attached to an ideal massless spring of spring constant 20 n/m along a horizontal, frictionless surface. the object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. (a) what is the amplitude of vibration? (b) at what location are the kineti

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Missing part in the question:
"B) At what location are the kinetic energy and the potential energy of the system the same?"

Solution)

A) amplitude of the vibration
The amplitude of the vibration corresponds to the maximum displacement of the mass-spring system with respect to its equilibrium position.

The energy of the mass-spring system at a generic point x (measured with respect to the equilibrium position) is sum of potential energy and kinetic energy:
$E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement
m is the mass of the object
v is the speed of the object when it is at position x

At the point of maximum displacement (i.e. when the displacement coincides to the amplitude, x=A), the speed of the object is zero, so the system has only potential energy:
$E= \frac{1}{2}kA^2$ (1)

Vice-versa, when the system crosses the equilibrium position, the displacement is zero (x=0) so the system has only kinetic energy, and the speed will be the maximum speed of the system:
$E= \frac{1}{2}mv_{max}^2$ (2)

The total energy must be conserved, so (1)=(2):
$\frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$
And since we know the spring constant, the mass of the object and its speed at the equilibrium position, we can find the amplitude A:
$A= \sqrt{ \frac{mv_{max}^2}{k} }= \sqrt{ \frac{(0.50 kg)(1.5 m/s)^2}{20 N/m} }=0.24 m$

B) We want to find the position x at which the kinetic energy and the potential energy are equal:
U=K

(3)
The only possibility to satisfy equation (3) is that both kinetic and potential energy are half of the total energy E:
$U=K= \frac{1}{2}E$
If we consider the potential energy only, this translates into
$\frac{1}{2}kx^2 = \frac{1}{2}E$ (4)

But the total energy E corresponds to the the potential energy of the system when it is at maximum displacement, x=A, so we can rewrite (4) as
$\frac{1}{2}kx^2 = \frac{1}{2}( \frac{1}{2}kA^2)$
which means
$x^2 = \frac{1}{2} A^2$
so we find
$x= \sqrt{ \frac{1}{2}A^2 }= \sqrt{ \frac{1}{2}(0.24 m)^2 }=0.17 m$
and this is the position at which the potential energy is equal to the kinetic energy of the system.

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A0.50-kg object is attached to an ideal massless spring of spring constant 20 n/m along a horizontal...