Is there a way to do repetitive tasks at intervals?

If you do not care about tick shifting (depending on how long did it took previously on each execution) and you do not want to use channels, it's possible to use native range function.

i.e.

package main

import "fmt"
import "time"

func main() {
    go heartBeat()
    time.Sleep(time.Second * 5)
}

func heartBeat() {
    for range time.Tick(time.Second * 1) {
        fmt.Println("Foo")
    }
}

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The function time.NewTicker makes a channel that sends a periodic message, and provides a way to stop it. Use it something like this (untested):

ticker := time.NewTicker(5 * time.Second)
quit := make(chan struct{})
go func() {
    for {
       select {
        case <- ticker.C:
            // do stuff
        case <- quit:
            ticker.Stop()
            return
        }
    }
 }()

You can stop the worker by closing the quit channel: close(quit).

Check out this library: https://github.com/robfig/cron

Example as below:

c := cron.New()
c.AddFunc("0 30 * * * *", func() { fmt.Println("Every hour on the half hour") })
c.AddFunc("@hourly",      func() { fmt.Println("Every hour") })
c.AddFunc("@every 1h30m", func() { fmt.Println("Every hour thirty") })
c.Start()

How about something like

package main

import (
    "fmt"
    "time"
)

func schedule(what func(), delay time.Duration) chan bool {
    stop := make(chan bool)

    go func() {
        for {
            what()
            select {
            case <-time.After(delay):
            case <-stop:
                return
            }
        }
    }()

    return stop
}

func main() {
    ping := func() { fmt.Println("#") }

    stop := schedule(ping, 5*time.Millisecond)
    time.Sleep(25 * time.Millisecond)
    stop <- true
    time.Sleep(25 * time.Millisecond)

    fmt.Println("Done")
}

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