Question

A block of unknown mass is attached to a spring of spring constant 7.3 N/m and undergoes simple harmonic motion with an amplitude of 12.7 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be 32.2 cm/s. Calculate the mass of the block. Answer in units of kg

Answer 1

Answer:

0.8516 kg

Explanation:

K = 7.3 N/m

A = 12.7 cm

y = A/2, v = 32.2 cm/s

Use the formula for the speed of mass

$v=\omega \sqrt{A^{2}-y^{2}}$

$32.2=\omega \sqrt{A^{2}-\frac{A^{2}}{4}}}$

$32.2=\omega \times \frac{\sqrt{3}}{2}\times 12.7$

ω = 2.93 rad/s

Now, $\omega =^{\sqrt{\frac{K}{m}}}$

Where, m is the mass of block.

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

m = 7.3 / (2.93 x 2.93) = 0.8516 kg