At $t=0$, a particle executing SHM with a time period $3 \mathrm{~s}$ is in phase with another particle…
At $t=0$, a particle executing SHM with a time period $3 \mathrm{~s}$ is in phase with another particle executing SHM. The time period of the second particle is $T$ (less than $3 \mathrm{~s}$ ). If they are again in the same phase for the third time after $45 \mathrm{~s}$, then the value of $T$ is .... .
1 s
1.5 s
2 s
2.5 s
Solution
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=0 \mathrm{~s}, 3 \mathrm{~s}, 6 \mathrm{~s}, 9 \mathrm{~s}, 12 \mathrm{~s}, 15 \mathrm{~s}, \ldots \text { 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 $t=15 \mathrm{~s}, 30 \mathrm{~s}, 45 \mathrm{~s} \ldots$ etc.
For second particle by option (a),
$
t=0 \mathrm{~s}, 1 \mathrm{~s}, 2 \mathrm{~s}, 3 \mathrm{~s}, 4 \mathrm{~s}, 5 \mathrm{~s}, 6 \mathrm{~s}, \ldots \text { etc }
$
By option (b),
$
t=0 \mathrm{~s}, 1.5 \mathrm{~s}, 3 \mathrm{~s}, 4.5 \mathrm{~s}, 6 \mathrm{~s}, \ldots \text { etc }
$
By option (c),
$
t=0 \mathrm{~s}, 2 \mathrm{~s}, 4 \mathrm{~s}, 6 \mathrm{~s}, \ldots \text { etc }
$
By option (d),
$
t=0 \mathrm{~s}, 25 \mathrm{~s}, 5 \mathrm{~s}, 7.5 \mathrm{~s}, 10 \mathrm{~s}, 125 \mathrm{~s}, 15 \mathrm{~s}, \ldots \text { etc }
$
So, both particles are in same phase again at $t=15 \mathrm{~s}, 30 \mathrm{~s}, 45 \mathrm{~s}, \ldots$ etc only, when time period of second particle is $2.5 \mathrm{~s}$