Answer:
Angular frequency will increase
No change in the amplitude
Explanation:
At extreme end of the SHM the energy of the SHM is given by

here we know that

now at the extreme end when one of the mass is removed from it
then in that case the angular frequency will change

So angular frequency will increase
but the position of extreme end will not change as it is given here that the top block is removed without disturbing the lower block
so here no change in the amplitude
The correct statements are that the speed decreases as the distance decreases and speed increases as the distance increases for the same time.
Answer:
Option A and Option B.
Explanation:
Speed is defined as the ratio of distance covered to the time taken to cover that distance. So Speed = Distance/Time. In other words, we can also state that speed is directly proportional to the distance for a constant time. Thus, the speed will be decreasing as there is decrease in distance for the same time. As well as there will be increase in speed as the distance increases for the same time. So option A and option B are the true options. So if there is decrease in the distance due to direct proportionality the speed will also be decreasing. Similarly, if the distance increases, the speed will also be increasing.
I believe the correct answer is C