Question

A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy of the system is 2.00 J. Find (a) the force constant of the spring and (b) the amplitude of the motion.

Answer 1

Answer:

(A) Spring constant will be 126.58 N/m

(B) Amplitude will be equal to 0.177 m

Explanation:

We have given mass of the block m = 200 gram = 0.2 kg

Time period T = 0.250 sec

Total energy is given TE = 2 J

(A) For mass spring system time period is equal to $T=2\pi \sqrt{\frac{m}{K}}$

So $0.250=2\times 3.14 \sqrt{\frac{0.2}{K}}$

$0.0398=\sqrt{\frac{0.2}{K}}$

Now squaring both side

$0.00158=\frac{0.2}{K}$

K = 126.58 N/m

So the spring constant of the spring will be 126.58 N/m

(B) Total energy is equal to $TE=\frac{1}{2}KA^2$, here K is spring constant and A is amplitude

So $2=\frac{1}{2}\times 126.58\times A^2$

$A^2=0.0316$

A = 0.177 m

So the amplitude of the wave will be equal to 0.177 m