Question

At $t=0$, a particle executing $SHM$ with a time period $3s$ is in phase with another particle executing SHM. The time period of the second particle is $T$ (less than $3s$ ). If they are again in the same phase for the third time after $45s$, then the value of $T$ is $\dots .$

• A. 1 s
• B. 1.5 s
• C. 2 s
• D. 2.5 s

Answer 1

Answer: (D)

Given, initially both oscillating particles are in same phase. For keeping calculation simple, we assume that they both are at mean positive $(x=0)$ at $t=0$
Now first particle is in same phase at given time instances.
$t=0s,3s,6s,9s,12s,15s,\dots$ etc.
Now, we take time period of second particle (from options) and select that value of $T$, which gives same phase of both particles at $t=15s,30s,45s\dots$ etc.
For second particle by option (a),
$t=0s,1s,2s,3s,4s,5s,6s,\dots$ etc
By option (b),
$t=0s,1.5s,3s,4.5s,6s,\dots$ etc
By option (c),
$t=0s,2s,4s,6s,\dots$ etc
By option (d),
$t=0s,25s,5s,7.5s,10s,125s,15s,\dots$ etc
So, both particles are in same phase again at $t=15s,30s,45s,\dots$ etc only, when time period of second particle is $2.5s$.