The frequency of oscillation on the frictionless floor is 28 Hz.
<h3>
Frequency of the simple harmonic motion</h3>
The frequency of the oscillation is calculated as follows;
f = (1/2π)(√k/m)
where;
- k is the spring constant
- m is mass of the block
f = (1/2π)(√7580/0.245)
f = 28 Hz
Thus, the frequency of oscillation on the frictionless floor is 28 Hz.
Learn more about frequency here: brainly.com/question/10728818
#SPJ1
Answer:
C)the right cord pulls on the pulley with greater force than the left cord
Explanation:
As we can see the figure that B is connected to the right string while A is connected to the left string
Now force equation for B as it will move downwards is given as

similarly for block A which will move upwards we can write the equation as

now we know that pulley is also rotating so the tangential acceleration of the rope at the contact point with pulley must be same as that of acceleration of the blocks
so here pulley will rotate clockwise direction
So tension in the right string must be more than the left string
So correct answer will be
C)the right cord pulls on the pulley with greater force than the left cord