Question

1) You put a 51.7 gram mass on a spring, set it in motion with a small amplitude, and count 21 cycles. Those 21 cycles took 3.42 seconds.
What is k_SHM?

2) Use Hooke's Law for this (F = - k s ): Where F is the spring's restoring force; k is the spring constant; and s is the stretch. The negative sign means the spring's restoring force is opposite the stretch direction.

You have a plot from weight [N] versus stretch [m]. The data forms a linear trend y = 3.662 * x + 1.67. How much will the spring stretch if 51.7 grams is hung on the spring?

Answer in centimeters with three significant figures or N/A if not enough information is given to answer. When you calculate your ansswer, don't use the negative sign in the Hooke's Law formula. Just know that the negative sign simply denotes the force direction is opposite the stretch (or compression).

Answer 1

1.

m = mass attached to spring = 51.7 gram = 0.0517 kg

n = number of cycles = 21

t = time taken to complete the cycles = 3.42 sec

Time period of each cycle is given as

T = t/n

T = 3.42/21 = 0.163 sec

k = spring constant

Time period of the simple harmonic motion of the mass attached is given as

T = 2 sqrt(m/k)

0.163 = (2 x 3.14) sqrt(0.0517/k)

k = 76.74 N/m

y = 3.662 x + 1.67

comparing the equation with general equation , y = mx + c

we get , m = slope = 3.662

we know that , slope of a force -stretch graph gives the force constant "k"

hence k = m = 3.662

m = mass attached = 51.7 g = 0.0517 kg

x = stretch of the spring

using equilibrium of force

mg = kx

(0.0517) (9.8) = (3.662) x

x = 0.14