a mass m at the end of a spring oscillates with a frequency of 0.83 hz. when an additional 780-g mass is added to m, the frequency is 0.60 hz. what is the value of m?



Answer :

Given the following data:

Initial frequency = 0.83 Hz.

Additional mass = 780 grams to kg = 0.78 kg

New frequency = 0.60 Hz

To determine the value of the initial mass, m, we would apply the following formula:

;f=1/2л⟌k/m

Where:

F is the frequency.

m is the mass

k is the spring constant.

For mass (m):

0.83=1/2л⟌k/m    ...equation 1.

For additional mass:

0.60=1/2⟌k/m+0.78  ...equation 2.

Next, we would divide eqn. 1 by eqn. 2:

;0.83/0.60=1/2л⟌((k/m)/1)/2л⟌m+0.78

1.3833=⟌k/m/⟌k/m+0.78

Taking the square of both sides, we have:

1.9135=k/m/k/m+0.78

1.9135=k/m×m+0.78/k

1.9135=m+0.78/m

1.9135m-m=0.78

0.9135m=0.78

m=0.78/0.9135

m=0.8538kg

the value of mass is 0.8538.

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