Question

A 0.273 kg body undergoes simple harmonic motion of amplitude 5.55 cm and period 0.250 s. (a) What is the magnitude of the maximum force acting on it? (b) If the oscillations are produced by a spring, what is the spring constant?

Answer 1

Answer:

Part a)

$F_{max} = 9.57 N$

Part b)

$k = 172.4 N/m$

Explanation:

Here we know that

Amplitude = 5.55 cm

Time period = 0.250 s

mass = 0.273 kg

Part a)

As we know that an object is executing SHM so the maximum acceleration of SHM is given as

$a_{max} = \omega^2 A$

$a_{max} = (\frac{2\pi}{0.250})^2(0.0555)$

$a_{max} = 35 m/s^2$

now we know that

$F_{max} = m a_{max}$

here we have

$F_{max} = 0.273(35) = 9.57 N$

Part b)

We know that angular frequency of SHM is given by following formula when it is a spring block system

$\omega^2 = \frac{k}{m}$

so here we have

$(\frac{2\pi}{T})^2 = \frac{k}{0.273}$

$(\frac{2\pi}{0.250})^2 = \frac{k}{0.273}$

$k = 172.4 N/m$