Question

A mass m is gently placed on the end of a freely hanging spring. The mass then falls 37 cm before it stops and begins to rise. What is the frequency of the oscillation?

Answer 1

We can solve this problem by using two formula of force which is F=kx ,Where,k=spring constant ,x=max distance spring compresses

F=mg , where m= mass, g= acceleration due to gravity and valued as 9.8m/s²

ω=√(k/m) ,where ω=angular frequency ,f=frequency of oscillation

Now we know that frequency is equal to

f=ω/2π

Now from above two formula of force we can say that

F=F

mg=kx

k=mg/x

Where we can find x by using, x =37/2=18.5cm= 0.185m

Now we will substitute the value of all the quantities in equation of frequency which is

f=ω/2π

f=√(k/m)/2π

f=√(mg/xm)/2π

f=√(g/x)/2π

f=√(9.8/.185)/2×3.14

f=1.157Hz