Question

A block of unknown mass is attached to a spring with a spring constant of 6.50 N/m and undergoes simple harmonic motion with an amplitude of 11.5 cm. When the block is halfway between its equilibrium position and the end point, its speed is measured to be 29.0 cm/s.
(a) Calculate the mass of the block.

(b) Calculate the period of the motion.

(c) Calculate the maximum acceleration of the block.

Answer 1

Explanation:

Given that,

Spring constant of the spring, k = 6.5 N/m

Amplitude of simple harmonic motion, A = 11.5 cm

Speed of the block, v = 29 cm/s = 0.29 m/s

(a) When the block is halfway between its equilibrium position and the end point, x = 5.75 cm

The speed of oscillator at a given position is given by :

$v=\sqrt{\dfrac{k}{m}(A^2-x^2)}$

$m=\dfrac{k}{v^2}(A^2-x^2)$

$m=\dfrac{6.5}{(0.29)^2}((11.5\times 10^{-2})^2-(5.75\times 10^{-2})^2)$

m = 0.766 kg

(b) The period of motion is given by :

$T=\dfrac{2\pi}{\omega}$

$\omega=\sqrt{\dfrac{k}{m}}$

$\omega=\sqrt{\dfrac{6.5}{0.766}}$

$\omega=2.91$

$T=\dfrac{2\pi}{2.91}$

T = 2.15 s

(c) The maximum acceleration of the block is given by :

$a=\omega^2A$

$a=(2.91)^2\times 11.5\times 10^{-2}$

$a=0.973\ m/s^2$

Hence, this is the required solution.