Question

When a 0.710 kg mass oscillates on an ideal spring, the frequency is 1.46 Hz .

A. What will the frequency be if 0.260 kg are added to the original mass? Try to solve this problem withoutfinding the force constant of the spring. B. What will the frequency be if 0.260 kg are subtracted from the original mass? Try to solve this problem without finding the force constant of the spring.

Answer 1

f = frequency = 1.46 Hz

m = mass = 0.71 kg

k = spring constant

frequency is given as

f = (1/(2)) sqrt(k/m₁)

1.46 = (1/6.28) sqrt(k/0.710)

k = 59.7 N/m

A)

m = new mass = 0.710 + 0.260 = 0.97 kg

f' = new frequency

using the formula

f' = (1/(2)) sqrt(k/m)

f' = (1/6.28) sqrt(59.7/0.97)

f' = 1.25 Hz

b)

m = new mass = 0.710 - 0.260 = 0.45 kg

f' = new frequency

using the formula

f' = (1/(2)) sqrt(k/m)

f' = (1/6.28) sqrt(59.7/0.45)

f' = 1.83 Hz

without calculating spring constant

m = mass = 0.71 kg

k = spring constant

frequency is given as

f = (1/(2)) sqrt(k/m)                      eq-1

A)

m' = new mass = m + 0.260

f' = new frequency

using the formula

f' = (1/(2)) sqrt(k/m')                                eq-2

dividing eq-1 by eq-2

f/f' = sqrt(m'/m)

1.46 /f' = sqrt((0.710 + 0.260) /0.710)

f' = 1.25 Hz

b)

m'' = new mass = 0.710 - 0.260 = 0.45 kg

f'' = new frequency

using the formula

f'' = (1/(2)) sqrt(k/m'')            eq-3

dividing eq-1 by eq-3

f/f'' = sqrt(m''/m)

1.46/f'' = sqrt(0.45/0.71)

f'' = 1.83 hz