Question

An oscillator consists of a block of mass 0.500 kg connected to a spring. When set into oscillation with amplitude 47.0 cm, it is observed to repeat its motion every 0.550 s. (a) Find the period. s (b) Find the frequency. hz (c) Find the angular frequency. rad/s (d) Find the spring constant. N/m (e) Find the maximum speed. m/s (f) Find the maximum force exerted on the block. N

Answer 1

a)
`T =0.55 s`

b)
`f =(1)/(T)=(1)/(0.55 s)=1.82 Hs`

c)
`w = 2 \pi *(1.82 Hz) =11.424 (rad)/(s)`

d)
`k = m (2\pi f)^2 = m w^2 = 0.5 kg * (11.424 rad/s)^2 =65.254 (N)/(m)`

e)
`v_(max)= Aw= 0.47 m * 11.424 (rad)/(s)=5.37 (m)/(s)`

f)
`F_(max)= A m w^2 = 0.47 m *0.5 kg* (11.424 (rad)/(s))^2 =30.67 N`

Step-by-step explanation:

Part a

For this case we know that the mass repeat its motion every 0.550 s. And that's the definition of period so for this case
`T =0.55 s`

Part b

By definition the frecuency is the inverse of the period so we have this:

`f =(1)/(T)=(1)/(0.55 s)=1.82 Hs`

Part c

The angular fecuency is defined with the following formula:

`w = 2\pi f`

And since we have the frequency we can replace:

`w = 2 \pi *(1.82 Hz) =11.424 (rad)/(s)`

Part d

For this case we know that the period is given by:

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

And if we solve for k we have this:

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

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

And since
`f=(1)/(T)` we can rewrite this expression like this:

`k = m (2\pi f)^2 = m w^2 = 0.5 kg * (11.424 rad/s)^2 =65.254 (N)/(m)`

Part e

The maximum speed for an oscillator is given by this formula:

`v_(max)= Aw= 0.47 m * 11.424 (rad)/(s)=5.37 (m)/(s)`

Part f

The maximum force is given by this formula:

`F_(max) = ma_(max)`

And the
`a_(max)= Aw^2` because the acceleration is the derivate of the velocity, so then we have:

`F_(max)= A m w^2 = 0.47 m *0.5 kg* (11.424 (rad)/(s))^2 =30.67 N`