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AQA A2 SHM quesitons
Hi everyone, just revising my A2 physics on SHM, and those 2 questions confused me, hope i can find some helps here:
Q1: An object oscillates in simple harmonic motion with an amplitude of 12mm and a time period of 0.27s.
Calculate: it's displacement and its direction of motion 0.20s after its displacement was +12mm
I've calculated it's displacement using formula X=ACOS(2*pi*f*t) and got answer -0.7mm, and i imagined when it's displacement was +12mm, it's at maximum positive displacement, and it would take 0.27*3/4 = 0.2025s to pass the equilibrium and return to the maximum positive placement, so i think the answer for this question should be -0.7mm towards equilibrium, but then i checked the answer from the book it says it's -0.7mm and towards maximum positive displacement...did the book just simply ignored the 0.2025 as 0.20 or did i get somewhere wrong?
Q2: A simple pendulum consists of a small weight on the end of a thread. The weight is displaced from equilibrium and released. It oscillates with an amplitude of 32mm, taking 20s to execute 10 oscillations, when the object is released at time t=0. State the displacement and calculate the acceleration when (i) t = 1.0s (ii) t= 1.5s
the answers in the book is (i) -32mm and 0.32ms-2 and (ii) 0,0 but i got the 0,0 results for (i) and -32mm for (ii), is this another error from the book?
Q1: An object oscillates in simple harmonic motion with an amplitude of 12mm and a time period of 0.27s.
Calculate: it's displacement and its direction of motion 0.20s after its displacement was +12mm
I've calculated it's displacement using formula X=ACOS(2*pi*f*t) and got answer -0.7mm, and i imagined when it's displacement was +12mm, it's at maximum positive displacement, and it would take 0.27*3/4 = 0.2025s to pass the equilibrium and return to the maximum positive placement, so i think the answer for this question should be -0.7mm towards equilibrium, but then i checked the answer from the book it says it's -0.7mm and towards maximum positive displacement...did the book just simply ignored the 0.2025 as 0.20 or did i get somewhere wrong?
Q2: A simple pendulum consists of a small weight on the end of a thread. The weight is displaced from equilibrium and released. It oscillates with an amplitude of 32mm, taking 20s to execute 10 oscillations, when the object is released at time t=0. State the displacement and calculate the acceleration when (i) t = 1.0s (ii) t= 1.5s
the answers in the book is (i) -32mm and 0.32ms-2 and (ii) 0,0 but i got the 0,0 results for (i) and -32mm for (ii), is this another error from the book?
Welcome to TSR.
1. You and the book are correct.
Yes they have rounded the amount (0.20s) to 2 sig figs because the data is given to 2 sig figs in the question.
The object has moved from max +ive displacement through equilibrium, out to max negative displacement and is now heading back towards max positive displacement. It has 0.7mm to go before it gets back to equilibrium. So it is heading both for equilibrium (almost there) and max positive displacement. It's just a case of how you express it.
2. The period is 2s (10 oscillations in 20s)
It starts from max displacement, so in 0.5s it's at equilibrium, at 1s it is at max negative displacement (half a cycle) at 1.5s it's at equilibrium again. At 2s it's back where it started.
You may be thinking of t=0s being at equilibrium. The question states t=0 is at max displacement.
1. You and the book are correct.
Yes they have rounded the amount (0.20s) to 2 sig figs because the data is given to 2 sig figs in the question.
The object has moved from max +ive displacement through equilibrium, out to max negative displacement and is now heading back towards max positive displacement. It has 0.7mm to go before it gets back to equilibrium. So it is heading both for equilibrium (almost there) and max positive displacement. It's just a case of how you express it.
2. The period is 2s (10 oscillations in 20s)
It starts from max displacement, so in 0.5s it's at equilibrium, at 1s it is at max negative displacement (half a cycle) at 1.5s it's at equilibrium again. At 2s it's back where it started.
You may be thinking of t=0s being at equilibrium. The question states t=0 is at max displacement.
(edited 12 years ago)
Original post by Zookinn
Hi everyone, just revising my A2 physics on SHM, and those 2 questions confused me, hope i can find some helps here:
Q1: An object oscillates in simple harmonic motion with an amplitude of 12mm and a time period of 0.27s.
Calculate: it's displacement and its direction of motion 0.20s after its displacement was +12mm
I've calculated it's displacement using formula X=ACOS(2*pi*f*t) and got answer -0.7mm, and i imagined when it's displacement was +12mm, it's at maximum positive displacement, and it would take 0.27*3/4 = 0.2025s to pass the equilibrium and return to the maximum positive placement, so i think the answer for this question should be -0.7mm towards equilibrium, but then i checked the answer from the book it says it's -0.7mm and towards maximum positive displacement...did the book just simply ignored the 0.2025 as 0.20 or did i get somewhere wrong?
Q2: A simple pendulum consists of a small weight on the end of a thread. The weight is displaced from equilibrium and released. It oscillates with an amplitude of 32mm, taking 20s to execute 10 oscillations, when the object is released at time t=0. State the displacement and calculate the acceleration when (i) t = 1.0s (ii) t= 1.5s
the answers in the book is (i) -32mm and 0.32ms-2 and (ii) 0,0 but i got the 0,0 results for (i) and -32mm for (ii), is this another error from the book?
Q1: An object oscillates in simple harmonic motion with an amplitude of 12mm and a time period of 0.27s.
Calculate: it's displacement and its direction of motion 0.20s after its displacement was +12mm
I've calculated it's displacement using formula X=ACOS(2*pi*f*t) and got answer -0.7mm, and i imagined when it's displacement was +12mm, it's at maximum positive displacement, and it would take 0.27*3/4 = 0.2025s to pass the equilibrium and return to the maximum positive placement, so i think the answer for this question should be -0.7mm towards equilibrium, but then i checked the answer from the book it says it's -0.7mm and towards maximum positive displacement...did the book just simply ignored the 0.2025 as 0.20 or did i get somewhere wrong?
Q2: A simple pendulum consists of a small weight on the end of a thread. The weight is displaced from equilibrium and released. It oscillates with an amplitude of 32mm, taking 20s to execute 10 oscillations, when the object is released at time t=0. State the displacement and calculate the acceleration when (i) t = 1.0s (ii) t= 1.5s
the answers in the book is (i) -32mm and 0.32ms-2 and (ii) 0,0 but i got the 0,0 results for (i) and -32mm for (ii), is this another error from the book?
The answer of -0.32ms-2 is correct this is the working: a= -(2pi/2)^2 * 0.032 = -0.315ms-2
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