At which angles is the tension in a pendulum string maximized and halved?

[3ephemeralwnd]

there is a pendulum with mass m, attached to the end of a string, oscillating back and forth between angles of -45 degrees and +45 degrees relative to the vertical axis

at which point would the tension be a maximum?
and at what 2 angles would the tension in the string be half of its maximum value?

Attempt:

i think the tension in the string is at maximum when the pendulum is at the very bottom, or when the angle is 0, because T = Fg at this point. (whereas everywhere else, the tension would only equal to the parallel component of the gravitational force, Fg(y))

so then i derived a formula for tension T = mgcos(theta)

if Tmax = mg , then half of the maximum tension must be 0.5 mg, correct?

So then, 0.5 mg = mg cos(theta)
0.5 = cos(theta)
and theta = 60 degrees

but i don't think that answer is correct because the range of angles according to the question is only between -45 and 45 degrees

any thoughts?

 

[tiny-tim]

hi 3ephemeralwnd!

you have to do F = ma …

what is "a" ?​

 

[3ephemeralwnd]

Fnet = fg(x)
ma = mgsin(theta)
a = gsin(theta) ?

 

[tiny-tim]

hi 3ephemeralwnd!

(just got up :zzz: …)

erm … i was expecting you to say that a is the https://www.physicsforums.com/library.php?do=view_item&itemid=27" v²/L plus the tangential acceleration dv/dt