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

Answer:

a) $T =0.55 s$

b) $f =\frac{1}{T}=\frac{1}{0.55 s}=1.82 Hs$

c) $w = 2 \pi *(1.82 Hz) =11.424 \frac{rad}{s}$

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

e) $v_{max}= Aw= 0.47 m * 11.424 \frac{rad}{s}=5.37 \frac{m}{s}$

f) $F_{max}= A m w^2 = 0.47 m *0.5 kg* (11.424 \frac{rad}{s})^2 =30.67 N$

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:

$f =\frac{1}{T}=\frac{1}{0.55 s}=1.82 Hs$

Part c

The angular fecuency is defined with the following formula:

$w = 2\pi f$

And since we have the frequency we can replace:

$w = 2 \pi *(1.82 Hz) =11.424 \frac{rad}{s}$

Part d

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

$T = 2\pi \sqrt{\frac{m}{k}}$

And if we solve for k we have this:

$(\frac{T}{2\pi})^2 = \frac{m}{k}$

$k = m (\frac{2\pi}{T})^2$

And since $f=\frac{1}{T}$ we can rewrite this expression like this:

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

Part e

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

$v_{max}= Aw= 0.47 m * 11.424 \frac{rad}{s}=5.37 \frac{m}{s}$

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:

$F_{max}= A m w^2 = 0.47 m *0.5 kg* (11.424 \frac{rad}{s})^2 =30.67 N$