Question

Two masses m1 and m2 are suspended together by a massless spring of spring constant k. when the masses are in equilibrium m1 is removed without disturbing the system, then the angular frequency and amplitude of oscillations of m2 are

Answer 1

Answer:

Angular Frequency: ω=$\sqrt{\frac{k}{m2} }$

Amplitude : y= A sin ($\sqrt{\frac{k}{m2} }$) t

Step-by-step explanation:

Mass 1 = m1

Mass 2= m2

Spring constant = k

Angular frequency = ?

amplitude = ?

Solution:

AS it is given that masses are in equilibrium so initially there is no motion

which means that there will be no angular velocity in the start

when m1 is removed equilibrium will be disturbed and an angular velocity is produced and it will revolve in a circle

We know the formula for angular frequency of an object

ω=$\sqrt{\frac{k}{m} }$

where k is spring constant and m is the mass of object

so for the given

angular frequency will be

ω=$\sqrt{\frac{k}{m2} }$

Now for amplitude we know the formula

y = A sin ω t

here t is the time period, A is amplitude and w is angular frequency

Putting the value of ω here

y= A sin ($\sqrt{\frac{k}{m2} }$) t