Question

g In a lab experiment, you are given a spring with a spring constant of 15 N/m. What mass would you suspend on the spring to have an oscillation period of 0.97 s when in SHM

Answer 1

Answer:

0.356kg

Explanation:

The period, T, of a mass on a spring of spring constant, k, in simple harmonic motion can be calculated as follows;

T = 2π $\sqrt{\frac{m}{k} }$         --------------------(i)

Where;

m = mass

From the question;

T = 0.97s

k = 15N/m

Taking π = 3.142 and substituting the values of T and k into equation (i) as follows;

0.97 = 2 x 3.142 x $\sqrt{\frac{m}{15} }$

0.97 = 6.284 x $\sqrt{\frac{m}{15} }$

$\frac{0.97}{6.284}$ = $\sqrt{\frac{m}{15} }$

0.154 = $\sqrt{\frac{m}{15} }$

Square both sides

0.154² = $\frac{m}{15}$

0.0237 = $\frac{m}{15}$

m = 0.356

Therefore, the mass that could be suspended on the spring to have an oscillation period of 0.97s when in SHM is 0.356kg

Answer 2

Answer:

0.358 kg

Explanation:

From simple harmonic motion,

T = 2π√(m/k)................ Equation 1

Where T = period of the spring, k = spring constant of the spring, m = mass suspended, π = pie

make m the subject of the equation

m = kT²/4π².................. Equation 2

Given: k = 15 N/m, T = 0.97 s, π = 3.14

Substitute into equation 2

m = 15(0.97²)/(4×3.14²)

m = 14.1135/39.4384

m = 0.358 kg.

Hence mass suspended = 0.358 kg