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Simple harmonic motion question help

hi so I was doing the summary questions for 18.3 and stuck on question 2 ( as follows)

the displacement of an object oscillating in simple harmonic motion varies with time according to the equation x (mm) = 12cos10t where t = time in seconds after the object was at max positive displacement.

so for a (i) it said to calculate amplitude which I got as 12mm.
a (ii) was to find time period which I got as 0.6283185307 a
question (b) states what will by the displacement of the object when t = 0.1

I put t = 0.1 in the equation and got 11.99817234 mm

but the answer at the back says 6.5 mm what's going on? could someone please help me on where I went wrong? thank you!

Reply 1

6.5 is obtained using radians instead of degrees. In SHM radians are usually used, but I think in the actual exam they will probably specify.

Reply 2

Original post by assassinjeev22
hi so I was doing the summary questions for 18.3 and stuck on question 2 ( as follows)

the displacement of an object oscillating in simple harmonic motion varies with time according to the equation x (mm) = 12cos10t where t = time in seconds after the object was at max positive displacement.

so for a (i) it said to calculate amplitude which I got as 12mm.
a (ii) was to find time period which I got as 0.6283185307 a
question (b) states what will by the displacement of the object when t = 0.1

I put t = 0.1 in the equation and got 11.99817234 mm

but the answer at the back says 6.5 mm what's going on? could someone please help me on where I went wrong? thank you!


cos10t is written in the form: cos(ωt)\text{cos}(\omega t)

where ω=2πf\omega = 2\pi f

i.e. 360o = 2π2\pi radians

Reply 3

Original post by assassinjeev22
hi so I was doing the summary questions for 18.3 and stuck on question 2 ( as follows)

the displacement of an object oscillating in simple harmonic motion varies with time according to the equation x (mm) = 12cos10t where t = time in seconds after the object was at max positive displacement.

so for a (i) it said to calculate amplitude which I got as 12mm.
a (ii) was to find time period which I got as 0.6283185307 a
question (b) states what will by the displacement of the object when t = 0.1

I put t = 0.1 in the equation and got 11.99817234 mm

but the answer at the back says 6.5 mm what's going on? could someone please help me on where I went wrong? thank you!


Convention is to work in radians for SHM rather than degrees - this would give 6.48... which rounds off to 6.5mm as required :smile:
(edited 7 years ago)

Reply 4

Original post by Protostar
Convention is to work in radians for SHM rather than degrees - this would give 6.48... which rounds off to 6.5mm as required :smile:


Oh my goodness thank you!

Reply 5

6.5 is obtained using radians instead of degrees. In SHM radians are usually used, but I think in the actual exam they will probably specify.


Awesome thank you!

Reply 6

Original post by uberteknik
cos10t is written in the form: cos(ωt)\text{cos}(\omega t)

where ω=2πf\omega = 2\pi f

i.e. 360o = 2π2\pi radians


Thank you!

Reply 7

Can you please do a full calculation of this because i have no idea how to make it into radians

Reply 8

Original post by assassinjeev22
hi so I was doing the summary questions for 18.3 and stuck on question 2 ( as follows)
the displacement of an object oscillating in simple harmonic motion varies with time according to the equation x (mm) = 12cos10t where t = time in seconds after the object was at max positive displacement.
so for a (i) it said to calculate amplitude which I got as 12mm.
a (ii) was to find time period which I got as 0.6283185307 a
question (b) states what will by the displacement of the object when t = 0.1
I put t = 0.1 in the equation and got 11.99817234 mm
but the answer at the back says 6.5 mm what's going on? could someone please help me on where I went wrong? thank you!

how did you get the amplitude

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