What is the period and frequency for \[\sin \left( 2\pi \dfrac{t}{5} \right)\]?

Hint: This type of question depends on the period and frequency of a sine function. Frequency is the number of complete cycles of waves passing a point in unit time. Period is the time taken by a complete cycle of the wave to pass a point. Hence frequency is the reciprocal of period. Hence, we can write it as, \[Frequency=\dfrac{1}{Period}\]. If T is the period of periodic function \[f\left( x \right)\] then \[f\left( x \right)=f\left( x+T \right)\]. Also we know that the \[\sin t\] is a periodic function with period \[2\pi \] relative to \[t\].

Complete step by step solution:
Now, we have to find the period and frequency for \[\sin \left( 2\pi \dfrac{t}{5} \right)\].
We know that \[\sin t\] is a periodic function with period \[2\pi \] relative to \[t\].
Hence, we can say that,
\[\Rightarrow \sin \left( 2\pi \dfrac{t}{5} \right)\] has a period of \[2\pi \] relative to \[2\pi \dfrac{t}{5}\].
\[\Rightarrow \sin \left( 2\pi \dfrac{t}{5} \right)\] has a period of \[\dfrac{2\pi }{\left( \dfrac{2\pi }{5} \right)}\] relative to \[\dfrac{2\pi \dfrac{t}{5}}{\left( \dfrac{2\pi }{5} \right)}\].
\[\Rightarrow \sin \left( 2\pi \dfrac{t}{5} \right)\] has a period of \[2\pi \times \dfrac{5}{2\pi }=5\] relative to \[2\pi \dfrac{t}{5}\times \dfrac{5}{2\pi }=t\]
\[\Rightarrow \sin \left( 2\pi \dfrac{t}{5} \right)\] has a period of \[5\] relative to \[t\].
Now, as we know that, Frequency is the number of complete cycles of waves passing a point in unit time. Period is the time taken by a complete cycle of the wave to pass a point. Hence frequency is the reciprocal of period.
\[\Rightarrow Frequency=\dfrac{1}{Period}\]
\[\Rightarrow Frequency=\dfrac{1}{5}\]
Thus, we can write,
Period of \[\sin \left( 2\pi \dfrac{t}{5} \right)=5\] and
Frequency of \[\sin \left( 2\pi \dfrac{t}{5} \right)=\dfrac{1}{5}\].

Note: In this type of question students may make mistakes in calculation of period. As we have to find the period relative to \[t\] and given function of sin is \[\sin \left( 2\pi \dfrac{t}{5} \right)\] , student have to divide \[2\pi \] by \[\dfrac{2\pi }{5}\] to obtain the value of period.