Question

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

• 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.
$t=0 s , 3 s , 6 s , 9 s , 1 2 s , 15 s , \ldots$ 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 $t=15 s , 30 s , 45 s \ldots$ etc.
For second particle by option (a),
$t=0 s , 1s , 2 s , 3 s , 4 s , 5 s , 6 s , \ldots $ etc
By option (b),
$t=0 s, 1.5 s, 3 s, 4.5 s, 6 s, \ldots $ etc
By option (c),
$t=0 s , 2 s , 4 s , 6 s , \ldots $ etc
By option (d),
$t=0 s , 25 s , 5 s , 7.5 s , 10 s , 125 s , 15 s , \ldots$ etc
So, both particles are in same phase again at $t=15 s , 30 s , 45 s , \ldots$ etc only, when time period of second particle is $ 2.5 s$.