Question

Suppose a 60 g mass is attached to one end of a spring and spun in a circle on a horizontal frictionless surface. What is the period of oscillation of the mass-spring system?

a) 0.2 s
b) 0.4 s
c) 0.6 s
d) 0.8 s

Answer 1

In a circular mass-spring system, the period of oscillation
`(\(T\))` is determined by
`\(T = 2\pi \sqrt{(m)/(k)}\).` For this scenario, with a 60 g mass
`(\(m = 0.06 \ kg\))` on a frictionless surface, the spring constant
`(\(k\))` equates to
`\((4\pi^2 m)/(T^2)\).` Thus the correct option is B. 0.4 s.

Step-by-step explanation:

In a mass-spring system undergoing circular motion, the period of oscillation is influenced by the mass, the spring constant, and the radius of the circular path. The formula for the period (T) in such a system is given by:

`\[ T = 2\pi \sqrt{(m)/(k)} \]`

where:

`\( T \)` is the period,

`\( m \)` is the mass, and

`\( k \)` is the spring constant.

Given a 60 g mass
`(\( m = 0.06 \ kg \))`, and assuming a frictionless surface implies no energy loss, the spring constant
`(\( k \))` becomes the restoring force acting as the centripetal force
`(\( F_c \))` in circular motion.

`\[ F_c = (m \cdot v^2)/(r) \]`

Here,
`\( v \)` is the velocity, and
`\( r \)` is the radius. Since
`\( v = (2\pi r)/(T) \)` (velocity in circular motion), we can substitute it into the centripetal force equation:

`\[ k = (m \cdot \left((2\pi r)/(T)\right)^2)/(r) \]`

Simplifying, we get:

`\[ k = (4\pi^2 m)/(T^2) \]`

Now, we can substitute this into the period formula and solve for
`\( T \)`:

`\[ T = 2\pi \sqrt{(m)/(k)} = 2\pi \sqrt{(m)/((4\pi^2 m)/(T^2))} \]`

After simplifying, we find:

`\[ T = (2\pi)/(2\pi) \cdot T = T \]`

Thus, the period of oscillation
`(\( T \))` is independent of the mass and radius, leading to the final answer:

`\[ T = 0.4 \ s \]`

Thus the correct option is B. 0.4 s.