Question

A block of unknown mass is attached to a spring of spring constant 9.4 N/m and undergoes simple harmonic motion with an amplitude of 15.5 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be 32.8 cm/s. Calculate the mass of the block. Answer in units of kg

Answer 1

Answer:

1.57 kg

Explanation:

Let the mass of block is m.

Spring constant, K = 9.4 N/m

Amplitude, A = 15.5 cm

y = A/2

v = 32.8 cm/s = 0.328 m/s

The velocity of the particle executing SHM is given by

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

where, ω is the angular frequency of SHM.

$0.328=\omega\sqrt{A^{2}-\left ( \frac{A}{2} \right )^{2}}$

$0.328=\omega\times 0.866 A$

0.328 = ω x 0.866 x 0.155

ω = 2.45 rad/s

Now the angular frequency is given by

$\omega = \sqrt\frac{K}{m}$

K = mω²

9.4 = m x 2.45 x 2.45

m = 1.57 kg