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The following question is about relation between the period and mass attached in spring.

The following question is about relation between the period and mass attached in spring-example-1
The following question is about relation between the period and mass attached in spring-example-1
The following question is about relation between the period and mass attached in spring-example-2
User Evanhaus
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1 Answer

13 votes
13 votes

ω = sqrt( k/m )

ω: angular frequency; m: mass; k: spring constant

ω = 2π/T

T: period (how long it takes for one revolution)

T = 4.59s/10rev = 0.459s/rev

ω = 2π/T = ω = 2π/0.459

ω = 13.6889 radians/s

One revolution, or 2π radians, is 8 cm. (4 cm to stretch out, the same 4 cm to retract)

13.6889 radians/s * (4cm/2πradians) = 86.01 cm/s

ω = 86.01 cm/s = 0.086 m/s

Let's take the first example, where mass is 50g and angular freq. is 4.59s.

50g = 0.05 kg (kilograms are standard in SI, not grams, and are used more directly in calculations relating to force)

ω = sqrt( k/m )

0.086 = sqrt( k/0.05 )

Square both sides

0.007396 = k/0.05

multiply both by 0.05

k = 0.0003698 N/m = 3.698 * 10^-4 N/m

Let's graph frequency in relation to mass (mass is the independent variable)

Let's calculate the slope of the period-to-mass graph

m (slope) = (T2 - T1)/(m2 - m1) (change in frequency over change in mass)

Let's use the second example for T2 and T2, and the first for ω1 and m1.

m = (6.32-4.59)/(100-50) = 1.73/50

m = 0.0346 seconds/gram, which means for every gram of mass added, the period for 10 revolutions increases by 0.0346 seconds.

User Sofi
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