Question

A block of mass 2M is attached to a massless spring with springÐconstant k. This block is connected to two other blocks of masses M and 2M using two
massless pulleys and strings. The accelerations of the blocks are a_, a_ and a_ as shown in the figure. the system is released from rest with the spring in its
unstretched state. The maximum extension of the spring is x_. Which of the following option(s) is/are correct?
[g is the acceleration due to gravity. neglect friction]
A. a_ Ð a_= a_ Ð a_
B. At an extension of X_/4 of the spring, the magnitude of acceleration of the block connected to the spring is 3g/10
C. X_= 4Mg
K

D. When spring achieves an extension of X_/2 for the first time, the speed of the block connected to the spring is 3g ÃM/5k.

Answer 1

Answer:

block connected to the spring is 3g/10

C. X_= 4Mg

Explanation:

When spring achieves an extension of X_/2 for the first time, the speed of the block connected to the spring is 3g ÃM/5k.

Answer 2

Explanation:

please coorporate with my handwriting

constraint of this system

a2-a1=a1-a3

Mg-T=Ma2-(1)

2Mg-T=2Ma3-(2)

2T-kx=2Ma1

from equations we get the values of a1,a2 and a3

a2=g-T/M

a3=g-T/2M

a1=T/M-kx/2M

sisbtitute these equations in constraint relation that is

2a1=a2+a3

2(T/M-kx/2M)=g-T/M+g-T/2M

so the value of T =4Mg/7+2kx/7

from third equation

a1=(4mg/7+2kx/7)/M+kx/2M

so

4g/7-3kx/14M=3k/14M*(x-4g/7*14M/4k)

a1=3k/14M*(x-8Mg/3k)

a1=-7W(omega)*2(x-xknot)-SHM eq(simple harmonic equation)

Mean position=x=xknot =8Mg/3k

Max extension x knot=16Mg/3k

V=omega*under rootA*2-x*2

x becomes 0 as know when it was in half extension then it was his mean position

so apply and calculate

if it is wrong in between or in explanation then i am very extremely sorry for that i am in class 9th i am learning this typeq of concept